Propositional natural deduction COMP2600 / COMP6260
Dirk Pattinson Australian National University
Semester 2, 2016 Major proof techniques
Three major styles of proof in logic and mathematics Model based computation: truth tables for propositional logic Algebraic proof: simplification rules e.g. De Morgan’s Deductive reasoning: rules of inference
I Natural deduction is one example
I Other examples: tableau calculi, resolution, Hilbert calculi
1 / 25 T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r))
2 / 25 T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T
2 / 25 T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T
2 / 25 T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T
2 / 25 F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T
2 / 25 F TF T F F F T F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T
2 / 25 F FT T F F T T F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T
2 / 25 F FF F F F F T
Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T
2 / 25 Example proof using truth tables
Statement to be proved: (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) For all 8 (= 23) possibilities of p,q,r, calculate truth value of the statement p q r q ∨ r p ∧ (q ∨ r) p ∧ q (p ∧ q) ∨ r (p ∧ (q ∨ r)) → ((p ∧ q) ∨ r)) T TT T T T T T T TF T T T T T T FT T T F T T T FF F F F F T F TT T F F T T F TF T F F F T F FT T F F T T F FF F F F F T
2 / 25 The notion of a deductive proof
A proof is a sequence of steps. Each step is either:
I an axiom or an assumption; or
I a statement which follows from previous steps via a valid rule of inference. Natural deduction:
I for each connective, there is an introduction and an elimination rule
I rules are formal, but resemble natural, i.e., human reasoning
3 / 25 Example of a natural deduction proof
Statement to be proved: (p ∧ (q ∨ r)) → ((q → s) ∨ p)
1 p ∧ (q ∨ r) Assumption
2 p ∧-E, 1
3 (q → s) ∨ p ∨-I, 2
4 (p ∧ (q ∨ r)) → ((q → s) ∨ p) →-I, 1–3
4 / 25 ∧-E (and elimination)
p ∧ q p ∧ q p q
Conjunction rules
∧-I (and introduction)
p q p ∧ q
5 / 25 Conjunction rules
∧-I (and introduction)
p q p ∧ q ∧-E (and elimination)
p ∧ q p ∧ q p q
5 / 25 Example
Commutativity of conjunction (derived rule)
p ∧ q q ∧ p
1 p ∧ q
2 p ∧-E, 1
3 q ∧-E, 1
4 q ∧ p ∧-I, 2, 3
6 / 25 →-E (implication elimination) p p → q q
Implication rules
→-I (implication introduction) [ p ] . q p → q This notation means that by assuming p, you can prove q. p is then discharged - no longer an assumption.
7 / 25 Implication rules
→-I (implication introduction) [ p ] . q p → q This notation means that by assuming p, you can prove q. p is then discharged - no longer an assumption.
→-E (implication elimination) p p → q q
7 / 25 1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
8 / 25 2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
8 / 25 3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
8 / 25 4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
3 p
8 / 25 5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
8 / 25 6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
8 / 25 Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
8 / 25 Example - transitivity of implication (derived rule)
p → q q → r We prove p → r
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 p → r →-I, 3–5
Lines 1 and 2 are the assumptions given, you may use these throughout the proof Line 3 is an assumption made within the proof, you may use it only within its scope (lines 3 to 5)
8 / 25 1 p → q
2 p
3 q →-E, 1, 2
4 p → q →-I, 2–3 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
Notation: justification of a step
9 / 25 2 p
3 q →-E, 1, 2
4 p → q →-I, 2–3 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
Notation: justification of a step
1 p → q
9 / 25 3 q →-E, 1, 2
4 p → q →-I, 2–3 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
Notation: justification of a step
1 p → q
2 p
9 / 25 4 p → q →-I, 2–3 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
Notation: justification of a step
1 p → q
2 p
3 q →-E, 1, 2
9 / 25 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
Notation: justification of a step
1 p → q
2 p
3 q →-E, 1, 2
4 p → q →-I, 2–3
9 / 25 Notation: justification of a step
1 p → q
2 p
3 q →-E, 1, 2
4 p → q →-I, 2–3 This is a rather silly proof, we succeed in proving what we started with. But it illustrates the meaning of the line number notation: →-E,1,2 means that rule →-E proves line 3 from lines 1 and2 →-I,2-3 means rule →-I proves line 4 from the fact that we could assume line 2 and (using that assumption) prove line 3
9 / 25 1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
10 / 25 2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
10 / 25 3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
2 q → r
10 / 25 4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
2 q → r
3 p
10 / 25 5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
10 / 25 6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
10 / 25 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
Rules involving assumptions
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5
10 / 25 Rules involving assumptions
1 p → q
2 q → r
3 p
4 q →-E, 1, 3
5 r →-E, 2, 4
6 q ∧ r WRONG ∧-I, 4, 5 If a statement is inside the scope of an assumption, then it depends on that assumption. Given p → q and q → r, we then assumed p and ”proved” q ∧ r, but q ∧ r depends on p. Indentation and vertical lines indicate scoping Similar to scoping in program code: eg lines 3 to 5 are a a method, and p is a local variable to that method.
10 / 25 1 p ∧ q
2 q ∧-E, 1
3 (p ∧ q) → q →-I, 1–2
1 p ∧ You are a giraffe
2 You are a giraffe ∧-E, 1
3 p ∧ You are a giraffe → You are a giraffe →-I, 1–2
Useless assumptions
You can assume anything, but it might not be useful.
11 / 25 2 q ∧-E, 1
3 (p ∧ q) → q →-I, 1–2
1 p ∧ You are a giraffe
2 You are a giraffe ∧-E, 1
3 p ∧ You are a giraffe → You are a giraffe →-I, 1–2
Useless assumptions
You can assume anything, but it might not be useful.
1 p ∧ q
11 / 25 3 (p ∧ q) → q →-I, 1–2
1 p ∧ You are a giraffe
2 You are a giraffe ∧-E, 1
3 p ∧ You are a giraffe → You are a giraffe →-I, 1–2
Useless assumptions
You can assume anything, but it might not be useful.
1 p ∧ q
2 q ∧-E, 1
11 / 25 1 p ∧ You are a giraffe
2 You are a giraffe ∧-E, 1
3 p ∧ You are a giraffe → You are a giraffe →-I, 1–2
Useless assumptions
You can assume anything, but it might not be useful.
1 p ∧ q
2 q ∧-E, 1
3 (p ∧ q) → q →-I, 1–2
11 / 25 Useless assumptions
You can assume anything, but it might not be useful.
1 p ∧ q
2 q ∧-E, 1
3 (p ∧ q) → q →-I, 1–2
1 p ∧ You are a giraffe
2 You are a giraffe ∧-E, 1
3 p ∧ You are a giraffe → You are a giraffe →-I, 1–2
11 / 25 ∨-E (or elimination)
[p][q] . . p ∨ q r r r
Disjunction rules
∨-I (or introduction)
p p p ∨ q q ∨ p
12 / 25 Disjunction rules
∨-I (or introduction)
p p p ∨ q q ∨ p
∨-E (or elimination)
[p][q] . . p ∨ q r r r
12 / 25 1 p ∨ q
2 p
. .
a r
b q
. .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
13 / 25 2 p
. .
a r
b q
. .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
13 / 25 . .
a r
b q
. .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
13 / 25 a r
b q
. .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
. .
13 / 25 b q
. .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
. .
a r
13 / 25 . .
c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
. .
a r
b q
13 / 25 c r
d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
. .
a r
b q
. .
13 / 25 d r ∨-E, 1, 2–a, b–c
∨-E template
1 p ∨ q
2 p
. .
a r
b q
. .
c r
13 / 25 ∨-E template
1 p ∨ q
2 p
. .
a r
b q
. .
c r
d r ∨-E, 1, 2–a, b–c
13 / 25 1 p ∨ q
2 p
3 q ∨ p ∨-I, 2
4 q
5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p
14 / 25 2 p
3 q ∨ p ∨-I, 2
4 q
5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
14 / 25 3 q ∨ p ∨-I, 2
4 q
5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
2 p
14 / 25 4 q
5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
2 p
3 q ∨ p ∨-I, 2
14 / 25 5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
2 p
3 q ∨ p ∨-I, 2
4 q
14 / 25 6 q ∨ p ∨-E, 1, 2–3, 4–5
Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
2 p
3 q ∨ p ∨-I, 2
4 q
5 q ∨ p ∨-I, 4
14 / 25 Example: commutativity of disjunction (derived rule)
p ∨ q q ∨ p 1 p ∨ q
2 p
3 q ∨ p ∨-I, 2
4 q
5 q ∨ p ∨-I, 4
6 q ∨ p ∨-E, 1, 2–3, 4–5
14 / 25 [¬p] ¬-E (not elimination) . (eliminates ¬ from q ∧ ¬q an assumption) p
Negation rules
Idea: assume the opposite of what you want to prove and find a contradiction — so your assumption must have been wrong
[p] ¬-I (not introduction) . q ∧ ¬q ¬p
15 / 25 Negation rules
Idea: assume the opposite of what you want to prove and find a contradiction — so your assumption must have been wrong
[p] ¬-I (not introduction) . q ∧ ¬q ¬p [¬p] ¬-E (not elimination) . (eliminates ¬ from q ∧ ¬q an assumption) p
15 / 25 1 p
2 ¬p
3 p ∧ ¬p ∧-I, 1, 2
4 ¬¬p ¬I, 2–3
Example: double negation introduction (derived rule)
p It is raining ¬¬p It is not the case that is is not raining
16 / 25 1 p
2 ¬p
3 p ∧ ¬p ∧-I, 1, 2
4 ¬¬p ¬I, 2–3
Example: double negation introduction (derived rule)
p It is raining ¬¬p It is not the case that is is not raining
16 / 25 2 ¬p
3 p ∧ ¬p ∧-I, 1, 2
4 ¬¬p ¬I, 2–3
Example: double negation introduction (derived rule)
p It is raining ¬¬p It is not the case that is is not raining
1 p
16 / 25 3 p ∧ ¬p ∧-I, 1, 2
4 ¬¬p ¬I, 2–3
Example: double negation introduction (derived rule)
p It is raining ¬¬p It is not the case that is is not raining
1 p
2 ¬p
16 / 25 Example: double negation introduction (derived rule)
p It is raining ¬¬p It is not the case that is is not raining
1 p
2 ¬p
3 p ∧ ¬p ∧-I, 1, 2
4 ¬¬p ¬I, 2–3
16 / 25 1 p ∧ ¬p
2 ¬q
3 p ∧ ¬p R, 1
4 q ¬E, 2–3
Example: contradiction elimination (derived rule)
p ∧ ¬p q
17 / 25 1 p ∧ ¬p
2 ¬q
3 p ∧ ¬p R, 1
4 q ¬E, 2–3
Example: contradiction elimination (derived rule)
p ∧ ¬p q
17 / 25 2 ¬q
3 p ∧ ¬p R, 1
4 q ¬E, 2–3
Example: contradiction elimination (derived rule)
p ∧ ¬p q
1 p ∧ ¬p
17 / 25 3 p ∧ ¬p R, 1
4 q ¬E, 2–3
Example: contradiction elimination (derived rule)
p ∧ ¬p q
1 p ∧ ¬p
2 ¬q
17 / 25 4 q ¬E, 2–3
Example: contradiction elimination (derived rule)
p ∧ ¬p q
1 p ∧ ¬p
2 ¬q
3 p ∧ ¬p R, 1
17 / 25 Example: contradiction elimination (derived rule)
p ∧ ¬p q
1 p ∧ ¬p
2 ¬q
3 p ∧ ¬p R, 1
4 q ¬E, 2–3
17 / 25 1 ¬¬p
2 ¬p
3 ¬p ∧ ¬¬p ∧-I, 1, 2
4 p ¬E, 2–3
Example: double negation elimination (derived rule)
¬¬p p
18 / 25 1 ¬¬p
2 ¬p
3 ¬p ∧ ¬¬p ∧-I, 1, 2
4 p ¬E, 2–3
Example: double negation elimination (derived rule)
¬¬p p
18 / 25 2 ¬p
3 ¬p ∧ ¬¬p ∧-I, 1, 2
4 p ¬E, 2–3
Example: double negation elimination (derived rule)
¬¬p p
1 ¬¬p
18 / 25 3 ¬p ∧ ¬¬p ∧-I, 1, 2
4 p ¬E, 2–3
Example: double negation elimination (derived rule)
¬¬p p
1 ¬¬p
2 ¬p
18 / 25 Example: double negation elimination (derived rule)
¬¬p p
1 ¬¬p
2 ¬p
3 ¬p ∧ ¬¬p ∧-I, 1, 2
4 p ¬E, 2–3
18 / 25 Equivalence
p ↔ q means p is true if and only if q is true We can make the definition p ↔ q ≡ (p → q) ∧ (q → p) which would naturally give us these rules introduction rule: p → q q → p p ↔ q elimination rules:
p ↔ q p ↔ q p → q q → p
19 / 25 Equivalence — Rules
Alternatively we can get rules which don’t involve the → symbol ↔-I (↔ introduction) [p][q] . . q p p ↔ q ↔-E (↔ elimination)
p ↔ q p p ↔ q q q p Note the similarities to the →-I and →-E rules
20 / 25 Which rule to use next?
Guided by the “form” of your goal, and what you already have proved “form” — ie, look at the connective: ∧,∨,→,¬ always can consider using ¬-E (not elimination) rule to prove p ∨ q, ∨-I (or introduction) may not work p q p ∨ q p ∨ q p may not be necessarily true, q may not be necessarily true
21 / 25 To prove p ∨ q, sometimes you need to do this:
1 Using ¬-E, assume ¬(p ∨ q) (hoping to prove some contradiction) 2 When is ¬(p ∨ q) true ? When both p and q false! 3 From ¬(p ∨ q) how to prove ¬p ? (next slide) 4 Having proved both ¬p and ¬q, prove some further contradiction ¬p → q Exercise: prove p ∨ q
22 / 25 1 ¬(p ∨ q)
2 p
3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
23 / 25 1 ¬(p ∨ q)
2 p
3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
23 / 25 2 p
3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
1 ¬(p ∨ q)
23 / 25 3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
1 ¬(p ∨ q)
2 p
23 / 25 4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
1 ¬(p ∨ q)
2 p
3 p ∨ q ∨-I, 2
23 / 25 5 ¬p ¬I, 2–4
Not-or elimination (derived rule)
¬(p ∨ q) ¬p
1 ¬(p ∨ q)
2 p
3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
23 / 25 Not-or elimination (derived rule)
¬(p ∨ q) ¬p
1 ¬(p ∨ q)
2 p
3 p ∨ q ∨-I, 2
4 (p ∨ q) ∧ ¬(p ∨ q) ∧-I, 1, 3
5 ¬p ¬I, 2–4
23 / 25 1 ¬q
2 p
3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
24 / 25 1 ¬q
2 p
3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
24 / 25 2 p
3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
1 ¬q
24 / 25 3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
1 ¬q
2 p
24 / 25 4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
1 ¬q
2 p
3 q your proof of q from p
24 / 25 5 ¬p ¬I, 2–4
Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
1 ¬q
2 p
3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
24 / 25 Proving a contrapositive rule
p In the same way, whenever you can prove any q ¬q then you can prove ¬p
1 ¬q
2 p
3 q your proof of q from p
4 q ∧ ¬q ∧-I, 1, 3
5 ¬p ¬I, 2–4
24 / 25 1 ¬(p ∨ ¬p)
2 ¬p ¬∨-E (previous slide), 1
3 ¬¬p ¬∨-E (previous slide), 1
4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
25 / 25 1 ¬(p ∨ ¬p)
2 ¬p ¬∨-E (previous slide), 1
3 ¬¬p ¬∨-E (previous slide), 1
4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
25 / 25 2 ¬p ¬∨-E (previous slide), 1
3 ¬¬p ¬∨-E (previous slide), 1
4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
1 ¬(p ∨ ¬p)
25 / 25 3 ¬¬p ¬∨-E (previous slide), 1
4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
1 ¬(p ∨ ¬p)
2 ¬p ¬∨-E (previous slide), 1
25 / 25 4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
1 ¬(p ∨ ¬p)
2 ¬p ¬∨-E (previous slide), 1
3 ¬¬p ¬∨-E (previous slide), 1
25 / 25 Law of the excluded middle (derived)
p ∨ ¬p “Everything must either be or not be.” – Russell
1 ¬(p ∨ ¬p)
2 ¬p ¬∨-E (previous slide), 1
3 ¬¬p ¬∨-E (previous slide), 1
4 ¬p ∧ ¬¬p ∧-I, 2, 3
5 p ∨ ¬p ¬E, 1–4
25 / 25