Sequent Introduction/Theorem Introduction 1. Quantifier Shift: In Predicate Logic, the following wff’s are logically equivalent: Quantifier Shift (QS) ; pg. 357 * Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) (ꓱv)Pv Δ (ꓯv)(Pv Δ) (ꓯv)Pv Δ (ꓱv)(Pv Δ) * Note: The only restriction is that the variable ‘v’ cannot appear freely anywhere in ‘Δ’. So, for instance, the following sequent is valid (from the right side of the 3rd row above): Example #1: Fa (ꓯx)Fx Ⱶ (ꓯx)(Fa Fx) 1 (1) Fa (ꓯx)Fx A ??? 1 (n) (ꓯx)(Fa Fx) ? Let’s start by assuming the antecedent of the conditional for the purpose of I: 1 (1) Fa (ꓯx)Fx A 2 (2) Fa Ass. (I) 1, 2 (3) (ꓯx)Fx 1, 2, E 1, 2 (4) Fx 3, ꓯE 1 (5) Fa Fx 2, 4, I 1 (6) (ꓯx)(Fa Fx) 5, ꓯI The move from (5) to (6) binds ‘x’ to ‘ꓯ’, but is permissible because it rests on a premise where ‘x’ was already bound. Let’s do another. Here’s one direction of the right side of the 4th row of QS, above: Example #2: (ꓱx)(Fx Ga) Ⱶ (ꓯx)Fx Ga 1 (1) (ꓱx)(Fx Ga) A ??? 1 (n) (ꓯx)Fx Ga ? 1 We’ll start by assuming the antecedent of the conclusion ‘(ꓯx)Fx’ for purposes of I. Then we’ll assume a free-variable version of (1) for purposes of ꓱE: 1 (1) (ꓱx)(Fx Ga) A 2 (2) (ꓯx)Fx Ass. (I) 3 (3) Fx Ga Ass. (ꓱE) 2 (4) Fx 2, ꓯE 2, 3 (5) Ga 3, 4, E 2 (6) (Fx Ga) Ga 3, 5, I 1, 2 (7) Ga 1, 6, ꓱE 1 (8) (ꓯx)Fx Ga 2, 7, I On our way to using ꓱE, we were able to obtain ‘Ga’ resting on lines (2) and (3), but we were able to obtain ‘Ga’ resting on lines (1) and (2) after performing ꓱE. Then it was a simple arrow-introduction, discharging our assumption on line (2) to get the conclusion. Similar derivations can be made for all of the QS equivalence relations, but now you are permitted to take a shortcut and introduce them all as sequent introductions. 2. Theorems: We can also prove new theorems in Predicate Logic that we previously could not prove in Propositional Logic. The claim that “All things are identical to themselves” is clearly true. It turns out it is a theorem, and the proof for it is simple: T34: Ⱶ (ꓯx)x=x We just introduce ‘x=x’ using =I, which doesn’t rest on any assumptions, and then we universalize it: - (1) x=x =I - (2) (ꓯx)x=x 1, ꓯI The conclusion (2) binds ‘x’ to ‘ꓯ’, but this is permissible even though ‘x’ is a free variable in line (1). Remember, that the restriction is that we cannot bind a variable using ꓯI if the universalized wff rests on an assumption where that variable is unbound (i.e., free). While it is true that (2) FOLLOWS FROM (1), it does not REST on (1)—note the dashes to the left indicating that (2) doesn’t rest on ANYTHING—and furthermore, (1) is not an assumption at all (hence there is no ‘1’ to the left and no ‘A’ to the right). One more: “If anything is a lion, then it is a lion.” This is a theorem too. T35: Ⱶ (ꓯx)(Lx Lx) 2 Here’s the proof: 1 (1) Lx Ass. (I) - (2) Lx Lx 1, 1, I - (3) (ꓯx)(Lx Lx) 2, ꓯI Again, (3) doesn’t rest on any ASSUMPTION where ‘x’ is unbound. It simply universalizes (2), which isn’t an assumption and doesn’t rest on ANYTHING. 3. Using Sequent Introduction: Consider: “The only friend I have is Elizabeth. Elizabeth is not Nancy. Nancy is a Canadian. Therefore, there is Canadian who is not my friend.” 1 (1) Fe (ꓯx)(Fx x=e) A 2 (2) e≠n A 3 (3) Cn A ??? 1, 2, 3 (n) (ꓱx)(Cx ¬Fx) ? Here, we can dig out the second conjunct of (1) using E and then perform ꓯE, replacing ‘x’ with ‘n’ since we know that ‘n’ is in the domain. Let’s do it the long way first: 1 (1) Fe (ꓯx)(Fx x=e) A 2 (2) e≠n A 3 (3) Cn A 1 (4) (ꓯx)(Fx x=e) 1, E 1 (5) Fn n=e 4, ꓯE 6 (6) n=e Ass. (Red. #1) - (7) n=n =I 6 (8) e=n 6, 7, =E 2, 6 (9) e=n e≠n 2, 8, I 2 (10) n≠e 6, 9, ¬I 11 (11) Fn Ass. (Red. #2) 1, 11 (12) n=e 5, 11, E 1, 2, 11 (13) n=e n≠e 10, 12, I 1, 2 (14) ¬Fn 11, 13, ¬I 1, 2, 3 (15) Cn ¬Fn 3, 14, I 1, 2, 3 (16) (ꓱx)(Cx ¬Fx) 15, ꓱI Notice all of the tedious steps that we had to take (in blue and green). The blue highlights all of the steps we had to take just to reverse ‘e≠n’ to ‘n≠e’. Let us call this ‘identity reversal’ (IDR). The green highlights a simple modus tollens (MT). 3 See how much simpler this derivation would be if we were permitted to introduce those two sequents along the way: 1 (1) Fe (ꓯx)(Fx x=e) A 2 (2) e≠n A 3 (3) Cn A 1 (4) (ꓯx)(Fx x=e) 1, E 1 (5) Fn n=e 4, ꓯE 2 (6) n≠e 2, SI (IDR) 1, 2 (7) ¬Fn 5, 6, SI (MT) 1, 2, 3 (8) Cn ¬Fn 3, 7, I 1, 2, 3 (9) (ꓱx)(Cx ¬Fx) 8, ꓱI So, just as in Propositional Logic, in Predicate Logic we are now permitted to introduce sequents into our derivations, simply noting ‘SI’ for ‘Sequent Introduction’ (or ‘TI’ for ‘Theorem Introduction’) and the name of the sequent or theorem being introduced. 4. Conclusion: All of the same sequents and theorems from Propositional Logic are valid in Predicate Logic as well. Here were some of them. (Keep in mind that ‘P’, ‘Q’, and ‘R’ may stand in for ANY well-formed formula) Some Sequents and Theorems from Unit Two Replacement Rules Distribution (Dist.) Common Argument Forms/Sequents P (Q R) (P Q) (P R) De Morgan’s Law (DeM) Modus Tollens (MT) ¬(P Q) ¬P ¬Q P (Q R) (P Q) (P R) P Q , ¬Q Ⱶ ¬P ¬(P Q) ¬P ¬Q Contraposition (Contra.) Hypothetical Syllogism (HS) P Q ¬(¬P ¬Q) P Q ¬Q ¬P P Q , Q R Ⱶ P R P Q ¬(¬P ¬Q) Exportation (Exp.) Disjunctive Syllogism (DS) Material Implication (Mat. Imp.) P (Q R) (P Q) R P Q , ¬P Ⱶ Q P Q ¬P Q P Q , ¬Q Ⱶ P Theorems Negative Implication (Neg. Imp.) ¬(P Q) P ¬Q The Law of Excluded Middle (LEM) Double Negation (DN) Ⱶ P ¬P P Ⱶ ¬¬P Association (Assoc.) P (Q R) (P Q) R The Law of Non-Contradiction Rabbits Out of Hats (Hat) P (Q R) (P Q) R (LNC) P , ¬P Ⱶ Q Ⱶ ¬(P ¬P) And for quick reference, here are some NEW sequents and theorems, which can now be introduced as well: 4 New Sequents and Theorems Replacement Rules Quantifier Shift (QS) ; pg. 357 Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) Δ (ꓱv)Pv (ꓱv)(Δ Pv) Δ (ꓯv)Pv (ꓯv)(Δ Pv) (ꓱv)Pv Δ (ꓯv)(Pv Δ) (ꓯv)Pv Δ (ꓱv)(Pv Δ) Quantifier Negation (QN) ; pg. 332 Identity Reversal (IDR) (ꓯx)Δx ¬(ꓱx)¬Δx a=b b=a (S224) (ꓱx)Δx ¬(ꓯx)¬Δx a≠b b≠a (lecture) Theorems Identity Generalization (ID Gen.) Conditional Generalization (Cond. Gen.) Ⱶ (ꓯx)x=x (T34) Ⱶ (ꓯx)(Δx Δx) (T35) 5 .