Proving Theorems
In unit one, we called a statement that is true on every interpretation a tautology. In this unit, we will call such a statement a theorem, where a theorem is a wff that relies on NO assumptions. We will now learn how to construct proofs for theorems.
1. Proving Theorems: The statement “¬¬P P” is a tautology. So, we should be able to construct a derivation of this statement which relies on ZERO assumptions. Such a derivation is called a proof. Here it is:
1 (1) ¬¬P A 1 (2) P 1, ¬E - (3) ¬¬P P 1, 2, I
Using arrow-introduction after deriving “P” from “¬¬P”, we have derived a statement in line (3) which relies on no assumptions. We have proved our first theorem:
T1: Ⱶ ¬¬P P
Let’s do another one:
T2: Ⱶ (¬Q ¬P) (P Q)
To prove this theorem, we need to show that the antecedent entails the consequent. So, we’ll need to assume the antecedent, and hope to use “I” in the final line. Like this:
1 (1) ¬Q ¬P A
???
? (n-1) P Q ? - (n) (¬Q ¬P) (P Q) ?, I
Because we want to derive “P Q” in line (n-1), we’ll want to assume its antecedent “P”, as well as the opposite of its consequent “¬Q” and hope for a contradiction. Like this:
1 (1) ¬Q ¬P A 2 (2) P Ass. (I) 3 (3) ¬Q Ass. (Red.)
???
? (n-1) P Q ? - (n) (¬Q ¬P) (P Q) ?, I
1 Let’s work through the steps to see where the derivation above leads:
1 (1) ¬Q ¬P A 2 (2) P Ass. (I) 3 (3) ¬Q Ass. (Red.) 1, 3 (4) ¬P 1, 3, E 1, 2, 3 (5) P ¬P 2, 4, I 1, 2 (6) ¬¬Q 3, 5, ¬I 1, 2 (7) Q 6, ¬E 1 (8) P Q 2, 7, I - (9) (¬Q ¬P) (P Q) 1, 8, I
Here, by using a reductio, we were able to show that P entails Q, so that we could derive “P Q” by “I”. Then, since we have shown that (1) entails (8), we were able to obtain the theorem in (9) by another “I”. Let’s do one more:
T3: Ⱶ P ¬P
This is known as the Law of Excluded Middle (LEM). Obviously, since this theorem is not a conditional, we’re not going to be able to use “I” to obtain it. So, we’re going to need to try for a reductio. Let’s begin by assuming the negation of the theorem, then:
1 (1) ¬(P ¬P) Ass. (Red.)
???
? (n-1) ¬¬(P ¬P) 1, ?, ¬I - (n) P ¬P ?, ¬E
It is not obvious how we are supposed to derive a contradiction from (1). It turns out that we need to ALSO assume “P” for a reductio, deriving a contradiction from “P” in order to obtain “¬P”. And THEN we can derive ANOTHER contradiction from “¬P” in order to obtain line (n-1). Like this:
1 (1) ¬(P ¬P) Ass. (Red.) 2 (2) P Ass. (Red.) 2 (3) P ¬P 2, I 1, 2 (4) (P ¬P) ¬(P ¬P) 1, 3, I 1 (5) ¬P 2, 4, ¬I (discharging line 2) 1 (6) P ¬P 5, I 1 (7) (P ¬P) ¬(P ¬P) 1, 6, I - (8) ¬¬(P ¬P) 1, 7, ¬I (discharging line 1) - (9) P ¬P 8, ¬E
2 2. Strategies for Proving Theorems: The route that we took to prove T3 was probably not very obvious. Here are some tips for proving theorems:
(a) If the main operator in the theorem to be proved is a “¬” or a “”, the proof will use a reductio. Assume the theorem’s opposite and try to derive a contradiction.
(b) If the main operator of the theorem to be proved is a “”, the proof will use arrow-introduction. Assume the antecedent of the theorem, and try to derive the theorem’s consequent.
(c) If the main operator of the theorem to be proved is a “”, the proof will use TWO arrow-introductions. One at a time, assume EACH of the two antecedents which compose the two conditionals of the bi-conditional, and try to derive the consequent of each.
(d) If the main operator of the theorem to be proved is a “”, then each conjunct is itself a theorem. One at a time, prove each of the conjunctions two conjuncts, and then use “I” at the end to obtain the target theorem.
3 Theorem Introduction
Knowing which statements are theorems can be helpful. We can make use of them in our other derivations as a sort of “shortcut”. For, we are now permitted to introduce any theorem that we have proved as an assumption in our derivations, which rests on NO other premises. For instance, let’s prove this sequent:
S86: P Q , ¬P R Ⱶ Q R
Now, it may not be immediately obvious how to obtain “Q R” from the premises, EXCEPT that we have just proved that “P ¬P” is a theorem. These two disjuncts just happen to be the antecedents of our two conditionals above. So, we can actually introduce that theorem as an assumption to help us. Like this:
1 (1) P Q A 2 (2) ¬P R A - (3) P ¬P TI (LEM) (T3 is the “Law of Excluded Middle”)
???
1, 2 (n) Q R ?
On line (3), we’ve introduce the theorem “T3”. We make a note of this by writing “TI” for “Theorem Introduction” and cite the number of the theorem. To the left, line (3) rests on NO assumptions, because T3 is a theorem. It turns out that we can assume each of the disjuncts of line (3), “P” and “¬P”, and see that they BOTH entail “Q R”. Like this:
1 (1) P Q A 2 (2) ¬P R A - (3) P ¬P TI (LEM) 4 (4) P Ass. (I) 1, 4 (5) Q 1, 4, E 1, 4 (6) Q R 5, I 1 (7) P (Q R) 4, 6, I 8 (8) ¬P Ass. (I) 2, 8 (9) R 2, 8, E 2, 8 (10) Q R 9, I 2 (11) ¬P (Q R) 8, 10, I 1, 2 (12) Q R 3, 7, 11, E
Here, we’ve introduced a theorem, and shown that—based on premises (1) and (2)— both disjuncts of the theorem (LEM) entail “Q R”. We then used disjunction- elimination to obtain the conclusion in line (12).
4 Sequent Introduction
We will also now be permitted to use a similar shortcut to introduce previously proved SEQUENTS rather than THEOREMS. To illustrate, let’s construct a derivation of the following sequent:
S87: ¬¬{[(P Q) R] S} Ⱶ [P (Q R)] T
It turns out that recalling a previous sequent that we already proved last week (in the lesson, “Derivations, part 2”) will be helpful. Here is that previous sequent:
S20: (P Q) R Ⱶ P (Q R) (This sequent is generally called “Exportation”)
The derivation we’re aiming for begins like this:
1 (1) ¬¬{[(P Q) R] S} A 1 (2) [(P Q) R] S 1, ¬E 1 (3) (P Q) R 2, E
???
1 (n) [P (Q R)] T ?
So far, we’ve just used “¬E” and “E” to obtain (3) from (1). Now look at sequent S20. Now look at the conclusion to be derived in line (n). Notice any similarities? Now, we COULD re-do ALL of the work that we did in our derivation of S20. But, as a shortcut, we can just skip straight to the conclusion of that derivation, and cite the sequent. Like this:
1 (1) ¬¬{[(P Q) R] S} A 1 (2) [(P Q) R] S 1, ¬E 1 (3) (P Q) R 2, E 1 (4) P (Q R) 3, SI (Exp) (S20 is called “Exportation”) 1 (5) [P (Q R)] T 4, I
On line (4), we just write “SI” for “Sequent Introduction” and cite the name or number of the sequent we’re making use of. This way, rather than re-going through all of the 6 lines that already went through to derive (4) from (3) when we were proving S20/Exp, we can just cite S20/Exp and be done with it, going straight from (3) to (4) in one line. Nice!
5 Substitution Instances
1. Theorem Introduction – Substitution Instances: When making use of theorem or sequent introduction, often the statements that you are using will not look EXACTLY like those of the theorem or sequent you want to use—though they will have the same FORM. For instance, it should be obvious that all of the following statements have the same form as “P ¬P”:
Original Statement P ¬P
New Statement Substitution Q ¬ Q P = Q (P Q) ¬(P Q) P = P Q (S T) ¬(S T) P = S T
It turns out that we are permitted to introduce ANY of the statements above as theorems which rely on no other premises. But, since they do not look exactly like “P ¬P”, we will introduce them as “TISI” for “Theorem Introduction – Substitution Instance” rather than simply “TI”. This is because each of the statements above takes “P ¬P” and substitutes in the wff’s listed under “substitution” for “P”.
For instance, take the following theorem, which we have just proved, above:
T2: Ⱶ (¬Q ¬P) (P Q)
Suppose that you want to introduce a theorem which has the same FORM as the above, but actually looks like this:
[¬(C D) ¬( A B)] [(A B) (C D)]
The statement above IS actually a substitution-instance of theorem T2. It has merely subbed in the following for “P” and “Q”:
Substitution P Q A B C D
So, if you want to introduce the statement above, you may introduce it as follows:
- (n) [¬(C D) ¬( A B)] [(A B) (C D)] TISI (T2)
6 2. Sequent Introduction – Substitution Instances: And you may now do the same for sequents. Consider:
S20: (P Q) R Ⱶ P (Q R) (Exp.)
Now, consider the following substitutions for “P”, “Q”, and “R”:
Substitution P Q R A B C D ¬E
If we made these substitutions, S20 would become the following:
[(A B) (C D)] ¬E Ⱶ (A B) [(C D) ¬E]
You now have the tools to introduce the sequent above, like this:
1 (m) [(A B) (C D)] ¬E ? 1 (n) (A B) [(C D) ¬E] m, SISI (Exp.)
Here, we’ve gone from line (m) to line (n) using a “Sequent Introduction – Substitution Instance”, which we abbreviate as “SISI”. Note also that line (n) relies on the same premise(s) that (m) relied on.
3. Examples: Let’s do a derivation that could benefit from TISI, and another that benefits from SISI. As an example of the first, let’s construct a derivation for:
S88: (¬¬R R) (¬S ¬T) Ⱶ T S
This derivation is going to make use of the following two theorems:
T1: Ⱶ ¬¬P P T2: Ⱶ (¬Q ¬P) (P Q)
The antecedent in the premise of S88 should look suspiciously like theorem T1. That is:
¬¬R R looks like ¬¬P P
Also, the consequent of the premise in S88 looks like the antecedent of T2, and the conclusion of the sequent S88 looks like the consequent of T2. That is:
¬S ¬T looks like ¬Q ¬P T S looks like P Q 7 So, our derivation will look like this:
1 (1) (¬¬R R) (¬S ¬T) A - (2) ¬¬R R TISI (T1) 1 (3) ¬S ¬T 1, 2, E - (4) (¬S ¬T) (T S) TISI (T2) 1 (5) T S 3, 4, E
This derivation could have been WAY longer, but by introducing substitution instances of theorems T1 and T2, it was only 5 lines. Neat!
Let’s do one that uses SISI:
S90: P Q , Q R , R S , S T , T U Ⱶ P U
This sequent will draw on that sequent which we called a hypothetical syllogism (HS):
S89: P Q , Q R Ⱶ P R
S90 is just an instance of repeatedly using S89 (HS). Here’s the derivation:
1 (1) P Q A 2 (2) Q R A 3 (3) R S A 4 (4) S T A 5 (5) T U A
???
1, 2, 3, 4, 5 (n) P U ?
Deriving “P R” from lines (1) and (2) is just a straightforward instance of using (HS), but all of the other derived wff’s will be substitution-instances of (HS). Like this:
1 (1) P Q A 2 (2) Q R A 3 (3) R S A 4 (4) S T A 5 (5) T U A 1, 2 (6) P R 1, 2, SI (HS) 1, 2, 3 (7) P S 3, 6, SISI (HS) 1, 2, 3, 4 (8) P T 4, 7, SISI (HS) 1, 2, 3, 4, 5 (9) P U 5, 8, SISI (HS)
8 Below are listed some of the more common sequent and theorem introductions. The replacement rules below are all of the form “Δ Ο”, where the “” symbol denotes logical equivalence. In this class, it will be permissible to replace “Δ” with “Ο” (or vice versa) during any derivation, and simply cite the rule used (e.g., write “SI (DeM)” for a sequent introduction, or “SISI (DeM)” if it is a substitution-instance of the introduced sequent). Theorems can be introduced into any derivation using “TI” or “TISI”. Finally, the conclusions of the arguments and sequents at the bottom can be introduced (using “SI” or “SISI”) whenever you already have all of the premises listed.
Theorems
The Law of Excluded Middle (LEM) The Law of Non-Contradiction (LNC) Ⱶ P ¬P (T3) Ⱶ ¬(P ¬P) (T4)
Replacement Rules
De Morgan’s Law (DeM) ¬(P Q) ¬P ¬Q (S70 ; S71) P Q ¬(¬P ¬Q) (S69 ; S99) ¬(P Q) ¬P ¬Q (S81 ; S82) P Q ¬(¬P ¬Q) (S62 ; S63)
Material Implication (Mat. Imp.) Negative Implication (Neg. Imp.) P Q ¬P Q (S72 ; S73) ¬(P Q) P ¬Q (S77 ; S78)
Association (Assoc.) Distribution (Dist.) P (Q R) (P Q) R (S6 ; S7) P (Q R) (P Q) (P R) (S43 ; S55) P (Q R) (P Q) R (S53 ; S54) P (Q R) (P Q) (P R) (S50 ; S51)
Contraposition (Contra.) Exportation (Exp.) P Q ¬Q ¬P (S65) P (Q R) (P Q) R (S20 ; S21)
Common Argument Forms/Sequents
Modus Tollens (MT) Hypothetical Syllogism (HS) P Q , ¬Q Ⱶ ¬P (S60) P Q , Q R Ⱶ P R (S89)
Disjunctive Syllogism (DS) P Q , ¬P Ⱶ Q (S94) P Q , ¬Q Ⱶ P (S95)
Rabbits Out of Hats (Hat) Double Negation (DN) P , ¬P Ⱶ Q (S93) P Ⱶ ¬¬P (S61)
Note: Introduced theorems will rely on NO assumptions (i.e., you will write ‘-‘ in the left column). Introduced replacements will rely on the same assumptions that the replaced formula relied on. When introducing a sequent, you will introduce the wff on the right side of the turnstile ‘Ⱶ’ of that sequent, and it will rely on whichever assumptions the wff’s on the left side of the turnstile ‘Ⱶ’ relied on.
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With the above theorems and sequents in mind, let’s do one more derivation:
¬(P ¬P) (¬Q ¬P) Ⱶ P Q
This one is similar to S88, above, and starts like this:
1 (1) ¬(P ¬P) (¬Q ¬P) A
???
1 (n) P Q ?
How to begin? Well, note that the antecedent of the conditional on line (1) just IS the Law of Excluded Middle. We are now permitted to introduce that antecedent as a theorem, and then use E on lines (1) and (2), as follows:
1 (1) ¬(P ¬P) (¬Q ¬P) A - (2) ¬(P ¬P) TI (LEM) 1 (3) ¬Q ¬P 1, 2, E
???
1 (n) P Q ?
Now what? Well, go scan the list of sequents and theorems again. Compare line (3) to each of the sequents that we are permitted to introduce. Notice anything? According to the ‘Contraposition’ sequent, line (3) is logically equivalent to our target on line (n)! So, we’re permitted to write ‘P Q’ on the very next line, citing Sequent Introduction (Contraposition), like this:
1 (1) ¬(P ¬P) (¬Q ¬P) A - (2) ¬(P ¬P) TI (LEM) 1 (3) ¬Q ¬P 1, 2, E 1 (4) P Q 3, SI (Contra.)
We’re done! See how much easier a derivation can be now that we’re allowed to use sequent and theorem introduction? Yay!
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