Mathematical Logic

Lecture 6: Soundness and Completeness of Tableaux and Resolution

Ashutosh Gupta

TIFR, India

Compile date: 2015-08-26

1 Where are we and where are we going?

We have seen

I propositional logic syntax and semantics

I normal forms

I proof methods tableaux and resolution We will see

I soundness and completeness of the proof systems

2 Topic 6.1

Soundness

3 Why soundness theorems?

We need to show that

if

our proof method proves a theorem

then

it is a valid formula in the logic.

4 Proof Hint. Induction hyp: F and ¬F has property Q. Now apply the original structural induction to complete this proof. Exercise 6.1 Complete the above proof.

Note: ∗Technically, only those formulas that do not contain ⊥, >,⊕ and ⇔.

Structural induction in uniform notation Since uniform notation absorbs single negations, original structural induction is not immediately applicable, we need the following principle of induction. Theorem 6.1 Every propositional formula∗ has a property Q if

I Base case: every atomic formula and its negation have property Q

I induction steps: if F has property Q so does ¬¬X, if α1 and α2 have property Q so does α, if β1 or β2 have property Q so does β,

5 Note: ∗Technically, only those formulas that do not contain ⊥, >,⊕ and ⇔.

Structural induction in uniform notation Since uniform notation absorbs single negations, original structural induction is not immediately applicable, we need the following principle of induction. Theorem 6.1 Every propositional formula∗ has a property Q if

I Base case: every atomic formula and its negation have property Q

I induction steps: if F has property Q so does ¬¬X, if α1 and α2 have property Q so does α, if β1 or β2 have property Q so does β,

Proof Hint. Induction hyp: F and ¬F has property Q. Now apply the original structural induction to complete this proof. Exercise 6.1 Complete the above proof.

5 Structural induction in uniform notation Since uniform notation absorbs single negations, original structural induction is not immediately applicable, we need the following principle of induction. Theorem 6.1 Every propositional formula∗ has a property Q if

I Base case: every atomic formula and its negation have property Q

I induction steps: if F has property Q so does ¬¬X, if α1 and α2 have property Q so does α, if β1 or β2 have property Q so does β,

Proof Hint. Induction hyp: F and ¬F has property Q. Now apply the original structural induction to complete this proof. Exercise 6.1 Complete the above proof.

Note: ∗Technically, only those formulas that do not contain ⊥, >,⊕ and ⇔. 5 Definition 6.2 A branch ρ of a tableaux is sat if the set of formulas that are labels of the nodes of the branch are satisfiable. If the model involved is m, we write m |= ρ. Definition 6.3 A tableaux T is sat if there is branch in the tableaux that is satisfiable. If the model involved is m, we write m |= T. Definition 6.4 A resolution derivation R is sat if the set of clauses in R is sat. If the model involved is m, we write m |= R.

Satisfiable tableaux/resolution derivation

Definition 6.1 (Recall) A set of formulas Σ is sat if there is a model m s.t. for each F ∈ Σ, m |= F. We write m |= Σ.

6 Definition 6.3 A tableaux T is sat if there is branch in the tableaux that is satisfiable. If the model involved is m, we write m |= T. Definition 6.4 A resolution derivation R is sat if the set of clauses in R is sat. If the model involved is m, we write m |= R.

Satisfiable tableaux/resolution derivation

Definition 6.1 (Recall) A set of formulas Σ is sat if there is a model m s.t. for each F ∈ Σ, m |= F. We write m |= Σ. Definition 6.2 A branch ρ of a tableaux is sat if the set of formulas that are labels of the nodes of the branch are satisfiable. If the model involved is m, we write m |= ρ.

6 Definition 6.4 A resolution derivation R is sat if the set of clauses in R is sat. If the model involved is m, we write m |= R.

Satisfiable tableaux/resolution derivation

Definition 6.1 (Recall) A set of formulas Σ is sat if there is a model m s.t. for each F ∈ Σ, m |= F. We write m |= Σ. Definition 6.2 A branch ρ of a tableaux is sat if the set of formulas that are labels of the nodes of the branch are satisfiable. If the model involved is m, we write m |= ρ. Definition 6.3 A tableaux T is sat if there is branch in the tableaux that is satisfiable. If the model involved is m, we write m |= T.

6 Satisfiable tableaux/resolution derivation

Definition 6.1 (Recall) A set of formulas Σ is sat if there is a model m s.t. for each F ∈ Σ, m |= F. We write m |= Σ. Definition 6.2 A branch ρ of a tableaux is sat if the set of formulas that are labels of the nodes of the branch are satisfiable. If the model involved is m, we write m |= ρ. Definition 6.3 A tableaux T is sat if there is branch in the tableaux that is satisfiable. If the model involved is m, we write m |= T. Definition 6.4 A resolution derivation R is sat if the set of clauses in R is sat. If the model involved is m, we write m |= R.

6 Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat

7 induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model.

7 Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ.

7 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule.

7 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T .

7 I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

7 Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ :

7 case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F .

7 Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2.

7 Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2.

7 case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2.

7 b. Prove if Σ is sat then a resolution derivation R for Σ is sat

Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof

7 Tableaux expansion preserves satisfiability Theorem 6.2 If Σ is sat then a tableaux T for Σ is sat Proof. Let model m |= Σ. base case: empty tableaux satisfies any model. induction step: Assume m |= T . Let ρ be a branch of T s.t. m |= ρ. Let T 0 be a tableaux obtained after application of an expansion rule. 0 0 0 I case ρ is not expanded in T : ρ is a branch of T and m |= T . 0 I case ρ is expanded using F ∈ Σ: m |= F , m |= ρF , and m |= T .

I case ρ is expanded using F ∈ ρ : Therefore, m |= F . case F = β : ρ is expanded into two branches ρβ1 and ρβ2. Due to semantics of β, m |= β1 or m |= β2. Therefore, m |= ρβ1 or m |= ρβ2. case F = α : .... case F = ¬¬G : .... Exercise 6.2 a. Complete the above proof b. Prove if Σ is sat then a resolution derivation R for Σ is sat 7 Proof. Let us suppose |= F does not hold. Therefore, for some model m, m |= ¬F . Therefore, there is no closed tableaux for {¬F }. Therefore, `pt F does not hold. Exercise 6.3 Prove if `pr F then |= F.

Tableaux method is sound

Theorem 6.3 If `pt F then |= F.

8 Therefore, for some model m, m |= ¬F . Therefore, there is no closed tableaux for {¬F }. Therefore, `pt F does not hold. Exercise 6.3 Prove if `pr F then |= F.

Tableaux method is sound

Theorem 6.3 If `pt F then |= F. Proof. Let us suppose |= F does not hold.

8 Therefore, there is no closed tableaux for {¬F }. Therefore, `pt F does not hold. Exercise 6.3 Prove if `pr F then |= F.

Tableaux method is sound

Theorem 6.3 If `pt F then |= F. Proof. Let us suppose |= F does not hold. Therefore, for some model m, m |= ¬F .

8 Therefore, `pt F does not hold. Exercise 6.3 Prove if `pr F then |= F.

Tableaux method is sound

Theorem 6.3 If `pt F then |= F. Proof. Let us suppose |= F does not hold. Therefore, for some model m, m |= ¬F . Therefore, there is no closed tableaux for {¬F }.

8 Exercise 6.3 Prove if `pr F then |= F.

Tableaux method is sound

Theorem 6.3 If `pt F then |= F. Proof. Let us suppose |= F does not hold. Therefore, for some model m, m |= ¬F . Therefore, there is no closed tableaux for {¬F }. Therefore, `pt F does not hold.

8 Tableaux method is sound

Theorem 6.3 If `pt F then |= F. Proof. Let us suppose |= F does not hold. Therefore, for some model m, m |= ¬F . Therefore, there is no closed tableaux for {¬F }. Therefore, `pt F does not hold. Exercise 6.3 Prove if `pr F then |= F.

8 Topic 6.2

Completeness

9 Stronger claim: finding the proof is decidable/semi-decidable

First we will see a general technique to prove completeness.

Completeness

The completeness property says that

if

there is a valid formula in the logic

then

there exists a proof in the proof system

10 First we will see a general technique to prove completeness.

Completeness

The completeness property says that

if

there is a valid formula in the logic

then

there exists a proof in the proof system

Stronger claim: finding the proof is decidable/semi-decidable

10 Completeness

The completeness property says that

if

there is a valid formula in the logic

then

there exists a proof in the proof system

Stronger claim: finding the proof is decidable/semi-decidable

First we will see a general technique to prove completeness.

10 Topic 6.3

Model existence theorem

11 Due to rules 2-4, if F is in M then some from sub(F ) must be in M. Hintikka sets may be obtained from Example 6.1 sets by downward saturation. {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))}

Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M

4. if β ∈ M then β1 ∈ M or β2 ∈ M

12 Due to rules 2-4, if F is in M then some from sub(F ) must be in M. Hintikka sets may be obtained from sets by downward saturation.

{(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))}

Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M

4. if β ∈ M then β1 ∈ M or β2 ∈ M

Example 6.1 {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set

12 Due to rules 2-4, if F is in M then some from sub(F ) must be in M. Hintikka sets may be obtained from sets by downward saturation.

{(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))}

Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M

4. if β ∈ M then β1 ∈ M or β2 ∈ M

Example 6.1 {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set

12 Due to rules 2-4, if F is in M then some from sub(F ) must be in M. Hintikka sets may be obtained from sets by downward saturation.

Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))}

Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M

4. if β ∈ M then β1 ∈ M or β2 ∈ M

Example 6.1 {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set

12 Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))}

Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M Due to rules 2-4, if F is in M then 4. if β ∈ M then β1 ∈ M or β2 ∈ M some from sub(F ) must be in M. Hintikka sets may be obtained from Example 6.1 sets by downward saturation. {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set

12 Hintikka set Definition 6.5 A set M of formulas is called Hintikka set if 1. for each p ∈ Vars, not both p ∈ M and ¬p ∈ M 2. if ¬¬F ∈ M then F ∈ M

3. if α ∈ M then α1 ∈ M and α2 ∈ M Due to rules 2-4, if F is in M then 4. if β ∈ M then β1 ∈ M or β2 ∈ M some from sub(F ) must be in M. Hintikka sets may be obtained from Example 6.1 sets by downward saturation. {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p)} is not a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬¬q, q} is a Hintikka set {(p ∧ (¬q ⇒ ¬p)), p, (¬q ⇒ ¬p), ¬p} is not a Hintikka set Exercise 6.4 Extend the following set into Hintikka sets

I {(p ∨ q), (¬p ∧ ¬q)}

I {¬(p ⇒ (q ⇒ p))}

I {¬(¬r ∨ (r ⇒ s)) ∨ (q ∧ (r ⇒ s))} 12 Proof. Hintikka sets explicate what syntax We construct a model m s.t. m |= M. can possibly say about semantics!! For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat

13 Hintikka sets explicate what syntax can possibly say about semantics!! For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M.

13 Hintikka sets explicate what syntax can possibly say about semantics!!

By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M. For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value.

13 Hintikka sets explicate what syntax can possibly say about semantics!!

base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M. For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F .

13 Hintikka sets explicate what syntax can possibly say about semantics!!

step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M. For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment

13 Hintikka sets explicate what syntax can possibly say about semantics!!

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M. For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

13 Hintikka sets explicate what syntax can possibly say about semantics!!

Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. We construct a model m s.t. m |= M. For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: ....

13 Exercise 6.5 Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M.

Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. Hintikka sets explicate what syntax We construct a model m s.t. m |= M. can possibly say about semantics!! For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: ....

13 Hintikka’s Theorem Theorem 6.4 Every Hintikka set M is sat Proof. Hintikka sets explicate what syntax We construct a model m s.t. m |= M. can possibly say about semantics!! For each p ∈ Vars, 1. if p ∈ M then m(p) := 1, 2. if ¬p ∈ M then m(p) := 0, and 3. otherwise assign m(p) any value. By the new structural induction we will show that for each F ∈ M, m |= F . base: trivially due to the assignment step: F ∈ M

I case F = ¬¬H: Since H ∈ M, m |= H. Therefore m |= ¬¬H

I case F = α: Since α1, α2 ∈ M, m |= α1 and m |= α2.Therefore, m |= α. I case F = β: .... Exercise 6.5

Show for a Hintikka set M, for each F either F 6∈ M or ¬F 6∈ M. 13 Note that the above definition defines a collection of sets The definition reads like Hinttika set but it is not.

Consistency property

Definition 6.6 Let C be a collection of sets of formulas. C is a consistency property if for each S ∈ C satisfies the following. 1. for each p ∈ Vars, either p ∈/ S or ¬p ∈/ S 2. if ¬¬F ∈ S then {F }∪S ∈ C

3. if α ∈ S then {α1, α2}∪S ∈ C

4. if β ∈ S then {β1}∪S ∈ C or {β2}∪S ∈ C

14 Consistency property

Definition 6.6 Let C be a collection of sets of formulas. C is a consistency property if for each S ∈ C satisfies the following. 1. for each p ∈ Vars, either p ∈/ S or ¬p ∈/ S 2. if ¬¬F ∈ S then {F }∪S ∈ C

3. if α ∈ S then {α1, α2}∪S ∈ C

4. if β ∈ S then {β1}∪S ∈ C or {β2}∪S ∈ C Note that the above definition defines a collection of sets The definition reads like Hinttika set but it is not.

14 Proof. Let C+ := {S0|S0 ⊆ S and S ∈ C}. We show C+ is consistency property. Consider S0 ∈ C+. By definition, there is S ∈ C s.t. S0 ⊆ S. 1. Therefore, S0 does not contain contradictory literals. 2. If ¬¬F ∈ S0. Therefore, ¬¬F ∈ S. Therefore, {F }∪S ∈ C. Therefore, {F }∪S0 ∈ C+. 3. ....

Exercise 6.6 Complete the above argument

Subset closed consistency property

Theorem 6.5 Every consistency property C can be extended to a consistency property that is subset closed.

15 Consider S0 ∈ C+. By definition, there is S ∈ C s.t. S0 ⊆ S. 1. Therefore, S0 does not contain contradictory literals. 2. If ¬¬F ∈ S0. Therefore, ¬¬F ∈ S. Therefore, {F }∪S ∈ C. Therefore, {F }∪S0 ∈ C+. 3. ....

Exercise 6.6 Complete the above argument

Subset closed consistency property

Theorem 6.5 Every consistency property C can be extended to a consistency property that is subset closed. Proof. Let C+ := {S0|S0 ⊆ S and S ∈ C}. We show C+ is consistency property.

15 1. Therefore, S0 does not contain contradictory literals. 2. If ¬¬F ∈ S0. Therefore, ¬¬F ∈ S. Therefore, {F }∪S ∈ C. Therefore, {F }∪S0 ∈ C+. 3. ....

Exercise 6.6 Complete the above argument

Subset closed consistency property

Theorem 6.5 Every consistency property C can be extended to a consistency property that is subset closed. Proof. Let C+ := {S0|S0 ⊆ S and S ∈ C}. We show C+ is consistency property. Consider S0 ∈ C+. By definition, there is S ∈ C s.t. S0 ⊆ S.

15 2. If ¬¬F ∈ S0. Therefore, ¬¬F ∈ S. Therefore, {F }∪S ∈ C. Therefore, {F }∪S0 ∈ C+. 3. ....

Exercise 6.6 Complete the above argument

Subset closed consistency property

Theorem 6.5 Every consistency property C can be extended to a consistency property that is subset closed. Proof. Let C+ := {S0|S0 ⊆ S and S ∈ C}. We show C+ is consistency property. Consider S0 ∈ C+. By definition, there is S ∈ C s.t. S0 ⊆ S. 1. Therefore, S0 does not contain contradictory literals.

15 Subset closed consistency property

Theorem 6.5 Every consistency property C can be extended to a consistency property that is subset closed. Proof. Let C+ := {S0|S0 ⊆ S and S ∈ C}. We show C+ is consistency property. Consider S0 ∈ C+. By definition, there is S ∈ C s.t. S0 ⊆ S. 1. Therefore, S0 does not contain contradictory literals. 2. If ¬¬F ∈ S0. Therefore, ¬¬F ∈ S. Therefore, {F }∪S ∈ C. Therefore, {F }∪S0 ∈ C+. 3. ....

Exercise 6.6 Complete the above argument

15 Theorem 6.6 if C is of finite character then C is subset closed Exercise 6.7 Prove the above theorem.

Finite character

Definition 6.7 A consistency property C has finite character if S ∈ C iff every finite subset of S is in C.

16 Finite character

Definition 6.7 A consistency property C has finite character if S ∈ C iff every finite subset of S is in C. Theorem 6.6 if C is of finite character then C is subset closed Exercise 6.7 Prove the above theorem.

16 Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character.

17 Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property.

17 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy.

17 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds.

17 Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0.

17 Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0.

17 Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C.

17 Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C.

17 Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C.

17 Therefore, S0 ∈ C+. 3. ....

Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C.

17 Extendable to finite character

Theorem 6.7 A subset closed consistency property C is extendable to one of finite character. Proof. claim: C+ := {S0|all finite subsets of S0are in C} is consistency property. Let S0 ∈ C+. We have four conditions to satisfy. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} ∈ C. contradiction. First cond. holds. 2. case ¬¬F ∈ S0: Consider finite set T ⊆ S0∪{F }. Therefore, (T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ⊆ S0. Therefore, {¬¬F }∪(T − {F }) ∈ C. Since C is consistency property, {¬¬F }∪(T − {F })∪{F } ∈ C. Therefore, {¬¬F }∪T ∪{F } ∈ C. Since C is subset closed, T ∈ C. Therefore, S0 ∈ C+. 3. ....

17 Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C.

18 Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

18 Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj .

18 Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ).

18 Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn

18 S Since C is of finite character, i Si ∈ C

Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C

18 Limits in finite character

Theorem 6.8 Let consistency property C is of finite character. If S1, S2,... is sequence of S members of C such that S1 ⊆ S2 ⊆ ... . Then, i Si ∈ C. Proof. S Consider finite set {F1,..., Fk } ⊆ i Si .

Let nj be the smallest number s.t. Fj ∈ Snj . Let n = max(n1,..., nk ). Therefore, {F1,..., Fk } ⊆ Sn Since C is subset closed, {F1,..., Fk } ∈ C S Since C is of finite character, i Si ∈ C

18 Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat.

19 Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?).

19 Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?).

19 S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise

19 Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn.

19 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C.

19 By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M.

19 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C.

19 Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction.

19 If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set.

19 Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M.

19 Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat.

Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly.

19 Model existence theorem Theorem 6.9 Let C be a consistency property. If S ∈ C, S is sat. Proof. Wlog, we assume C is of finite character (why?). Let F1, F2,... be enumeration of all the formulas in some order(why?). Let us define a sequence S1, S2,... as follows. ( Sn∪{Fn} Sn∪{Fn} ∈ C S1 = SSn+1 = Sn otherwise S S Since Sn are in C and C is of finite character, nSn ∈ C. Let M := nSn. Claim: M is maximal in C. 0 0 0 Assume M ∈ C s.t.M ⊂ M . There is Fn such that Fn ∈ M and Fn ∈/ M. By def. of M, Sn∪{Fn} ∈/ C. 0 Since Sn∪{Fn} ⊆ M and C is subset closed, Sn∪{Fn} ∈ C. Contradiction. Claim: M is a Hinttika set. If α ∈ M then {α1, α2}∪M ∈ C. Since M is maximal, {α1, α2} ⊆ M. Other conditions holds similarly. Since M is a Hinttika set, M is sat. Since S ⊆ M, S is sat. 19 Evidence of unsatisfiablity is always a finite subset.

Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }.

20 Evidence of unsatisfiablity is always a finite subset.

Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property.

20 Evidence of unsatisfiablity is always a finite subset.

1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions.

20 Evidence of unsatisfiablity is always a finite subset.

0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds.

20 Evidence of unsatisfiablity is always a finite subset.

0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S .

20 Evidence of unsatisfiablity is always a finite subset.

Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T .

20 Evidence of unsatisfiablity is always a finite subset.

0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat.

20 Evidence of unsatisfiablity is always a finite subset.

0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat.

20 Evidence of unsatisfiablity is always a finite subset.

0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat.

20 Evidence of unsatisfiablity is always a finite subset.

Due to model existence theorem, S is sat.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven.

20 Evidence of unsatisfiablity is always a finite subset.

Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Proof. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

20 Compactness theorem

Theorem 6.10 If every finite subset of S is sat then S is sat Evidence of unsatisfiablity Proof. is always a finite subset. Let C := {S0|all finite subsets of S0 are sat }. Claim: C is a consistency property. Let S0 ∈ C. We need to satisfy the four conditions. 1. If {p, ¬p} ⊆ S0, then {p, ¬p} is sat. contradiction. First cond. holds. 0 0 2. Let α ∈ S . Consider a finite T ⊆ {α1, α2}∪S . 0 0 0 There is a finite T ⊆ S s.t. T ⊆ {α, α1, α2}∪T . Since T 0∪{α} ⊆ S0, T 0∪{α} is sat. 0 Therefore, T ∪{α, α1, α2} is sat. 0 Therefore, every finite subset of {α1, α2}∪S is sat. 0 Therefore, {α1, α2}∪S ∈ C 3. .... similarly other cases are proven. Due to model existence theorem, S is sat.

20 Topic 6.4

Completeness of Tableaux and Resolution

21 Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property.

22 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold.

22 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ

22 Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T .

22 Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction.

22 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof)

Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set.

22 Tableaux completeness Theorem 6.11 The collection of sets of formulas that are tableaux consistent is a consistency property. Proof. Let Σ be a tableaux consistent set. We need to show the 4 conditions hold. 1. If {p, ¬p} ⊆ Σ then there is a closed tableaux. Therefore, {p, ¬p} 6⊆ Σ 2. If ¬¬F ⊆ Σ. Suppose {F }∪Σ has a closed tableaux T . Then, we can construct a closed tableaux for Σ as follows. ¬¬F

F

T 00 where T 00 is obtained by removing all the nodes with label F in T if the node was added due to the introduction rule. Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. ... other cases have similar proofs. Exercise 6.8 (Complete the above proof) 22 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T . Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. case α ⊆ Σ: similarly as above

Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set.

23 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T . Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. case α ⊆ Σ: similarly as above

Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set. 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ

23 Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. case α ⊆ Σ: similarly as above

Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set. 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T .

23 Therefore, {F }∪Σ is a tableaux consistent set. 3. case α ⊆ Σ: similarly as above

Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set. 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T . Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction.

23 3. case α ⊆ Σ: similarly as above

Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set. 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T . Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction. Therefore, {F }∪Σ is a tableaux consistent set.

23 Resolution completeness Theorem 6.12 The collection of sets of formulas that are resolution consistent is a consistency property. Proof. Let Σ be a resolution consistent set. 1. If {p, ¬p} ⊆ Σ then there is a closed derivation. Therefore, {p, ¬p} 6⊆ Σ 2. case ¬¬F ⊆ Σ: Suppose {F }∪Σ has a closed derivation T . Then, we can construct a closed derivation for Σ as follows. {¬¬F } {F } T 0 where T 0 is obtained by deleting occurrences of {F } from T . Contradiction. Therefore, {F }∪Σ is a tableaux consistent set. 3. case α ⊆ Σ: similarly as above 23 Now constructing a closed derivation for Σ is not straight forward.

Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Resolution completeness (contd.) Proof(contd.) 4. case β ⊆ Σ:

24 Now constructing a closed derivation for Σ is not straight forward.

0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Resolution completeness (contd.) Proof(contd.) 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2.

24 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2.

24 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1

24 If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows.

24 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?).

24 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1

24 0 case T1 has {β2}: 0 T2 := remove {β2} in T2. {β} {β1, β2} 0 T1 0 T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 case T1 has {}: {β} {β1, β2} 0 T1 is closed derivation of Σ. contradiction

24 {β1}∪Σ or {β2}∪Σ is resolution consistent.

Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

24 Now constructing a closed Resolution completeness (contd.) derivation for Σ is not straight Proof(contd.) forward. 4. case β ⊆ Σ: Suppose both {β1}∪Σ and {β2}∪Σ have closed derivations T1 and T2. 0 We define derivation T1 as follows: 0 I T1 := replace {β1} clauses by {β1, β2} in T1 0 0 0 I Repeat i ∈ 1..|T1|, T1 := repair ith clause C in T1 as follows. If any antecedent of C is extended by β2 then apply the expansion rule again and obtain a replacement, which is either C or C∪{β2}(why?). 0 I T1 := remove {β1, β2} in T1 0 0 case T1 has {}: case T1 has {β2}: 0 {β} T2 := remove {β2} in T2. {β1, β2} {β} 0 T1 {β1, β2} 0 is closed derivation of Σ. T1 0 contradiction T2 is closed derivation of Σ. contradiction

{β1}∪Σ or {β2}∪Σ is resolution consistent. 24 Topic 6.5

Proofs are Enumerable

25 Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable.

26 Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables.

26 Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable.

26 Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F .

26 Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ.

26 There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn}

26 Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?).

26 We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F .

26 Therefore, eventually we will say yes if Σ |= F .

Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable.

26 Implication is effectively enumerable. Theorem 6.13 If Σ is a finite set of formulas, then Σ |= F is decidable. Proof. Due to truth tables. Theorem 6.14 If Σ is effectively enumerable, then Σ |= F is semi-decidable. Proof. Due to compactness theorem, if Σ |= F there is a finite set Σ0 such that Σ0 |= F . Since Σ is effectively enumerable, let G1, G2, .... be the enumeration of Σ. Let Sn := {G1,..., Gn} There must be a Σ0 ⊆ Sk (why?). Therefore, Sk |= F . We may enumerate Sn and check Sn |= F , which is decidable. Therefore, eventually we will say yes if Σ |= F . 26 End of Lecture 6

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