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Proving Theorems 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.
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