: Natural Deduction 26 October 2011 Notes on the sequent calculus proofs One of the so-called structural rules we didn t discuss in class is the weakening rule. The weakening rule (WEAK) states that we can always add new assumptions to our context, provided the new variable does not occur free in the λ-term (aka, does not clash with a variable of the same name in the context). The idea is that if φ is true under a certain set of assumptions, then it is true under an even greater set of assumptions. This is our metatheoretic version of Strengthening the Antecedent. We ll assume that the premises (h 1,h 2,...) are given their proper types in Γ. We want our proofs to be parametric in in the context so that people using these proofs don t have to worry about weakening them before they can use them, so we ll pass around an arbitrary Γ instead of hard-coding Γ=. And finally, for sequent calculus I m just going to assume the definition φ def = φ, and so I ll be using I and instead of I and E. 1 Alternative rules 1.1 Modus Tollens (MT): P Q, Q P Natural deduction P Qtrue Q true P true u true P true Q true E I u 26 October 2011 1 of 18
Fitch-style 1 P Qtrue Premise 2 Q true Premise 3 P true Assumption 4 Q true 1, 3, 5 true 2, 4, E 6 P true 3 5, I Sequent calculus N.B., this is the same as the hypothetical syllogism with R=. Γ h 1 :P Q Γ, u:p h 1 :P Q WEAK Γ, u:p u:p VAR Γ, u:p h 1 u : Q Γ, u:p h 2 (h 1 u) : Γ λu. h 2 (h 1 u) : P Γ h 2 :Q Γ, u:p h 2 :Q WEAK I 1.2 Disjunctive Syllogism (DS): P Q, P Q Naturally, we ll also have a proof of P Q, Q P. Rather than using the commutativity of disjunction followed by DS, we could prove it directly using the same idea as we have here. Natural deduction, with the implicational E P Q true P true u P true E true Q true P Qtrue Q true E I u Q true v Q Qtrue Iv E Natural deduction with ATP s E and E P Q true P true u P true E Q true Q true Q true v E u,v 26 October 2011 2 of 18
Fitch-style 1 P Q true Premise 2 P true Premise 3 P true Assumption 4 true 2, 3, E 5 Q true 4, E 6 P Qtrue 3 5, I 7 Q true Assumption 8 Q true 7 9 Q Qtrue 7 8, I 10 Q true 1, 6, 9, E Sequent calculus, with the implicational E Γ h 1 :P Q Γ h 2 :P Γ, u:p u:p VAR Γ, u:p h 2 :P WEAK Γ, u:p h 2 u : E Γ, u:p abort(h 2 u) : Q I Γ λu. abort(h 2 u) : P Q Γ either(λu. abort(h 2 u))(λv. v) h 1 : Q Γ, v:q v:q VAR Γ λv. v : Q Q I E Sequent calculus, with ATP s E Γ h 1 :P Q Γ h 2 :P Γ, u:p u:p VAR Γ, u:p h 2 :P WEAK Γ, u:p h 2 u : E Γ, u:p abort(h 2 u) : Q Γ case h 1 of inl u abort(h 2 u) inr v v:q Γ, v:q v:q VAR E u,v 1.3 Hypothetical Syllogism (HS): P Q,Q R P R Natural deduction P true u P Qtrue Q true R true P Rtrue Q Rtrue I u 26 October 2011 3 of 18
Fitch-style 1 P Qtrue Premise 2 Q Rtrue Premise 3 P true Assumption 4 Q true 1, 3, 5 R true 2, 4, 6 P Rtrue 3 5, I Sequent calculus N.B., this is just function composition. Γ h 1 :P Q Γ, u:p u:p VAR Γ, u:p h 1 :P Q WEAK Γ, u:p h 1 u : Q Γ, u:p h 2 (h 1 u) : R Γ λu. h 2 (h 1 u) : P R Γ h 2 :Q R Γ, u:p h 2 :Q R WEAK I 1.4 Constructive Dilemma (CD): P Q,R S,P R Q S We could reuse WC (proof 1.7) for half of the work here, but we d need the commuted variation in order to do the other half of the work before putting things together. So, for the sake of uniformity, we re just proving everything directly rather than reusing WC. Natural deduction, with ATP s E P R true P Qtrue P true u Q true I L Q S true Q S true R Strue R true v S true Q S true I R E u,v 26 October 2011 4 of 18
Fitch-style 1 P Qtrue Premise 2 R Strue Premise 3 P R true Premise 4 P true Assumption 5 Q true 1, 4, 6 Q S true 5, I L 7 P Q S true 4 6, I 8 R true Assumption 9 S true 2, 8, 10 Q S true 9, I R 11 R Q S true 8 10, I 12 Q S true 3, 7, 11, E Sequent calculus, with ATP s E Let the following derivation be called D: Γ h 1 : P Q Γ, u:p h 1 : P Q WEAK Γ, u:p u:p VAR Γ, u:p h 1 u : Q Γ, u:p inl(h 1 u) : Q S I L And let the following derivation be called E : Γ h 2 : R S Γ, v:r h 2 : R S WEAK Γ, v:r v:r VAR Γ, v:p h 2 v : S Γ, v:p inr(h 2 v) : Q S I R And finally we put those together with ATP-style disjunction elimination. 26 October 2011 5 of 18
Γ h 3 : P R Γ h 1 : P Q D Γ, u:p inl(h 1 u) : Q S Γ h 2 : R S E Γ, v:p inr(h 2 v) : Q S Γ case h 3 of inl u inl(h 1 u) inr v inr(h 2 v) : Q S E u,v 1.5 Destructive Dilemma (DD): P Q,R S, Q S P R This is like the constructive dilemma, except replacing the use of modus ponens with modus tollens. Natural deduction, with ATP s E Q S true P Qtrue Q true u P true P R true proof 1.1 I L P R true R Strue S true v R true P R true proof 1.1 I R E u,v Fitch-style 1 P Qtrue Premise 2 R Strue Premise 3 Q S true Premise 4 Q true Assumption 5 P true 1, 4, MT (proof 1.1) 6 P R true 5, I L 7 Q P R true 4 6, I 8 S true Assumption 9 R true 2, 8, MT (proof 1.1) 10 P R true 9, I R 11 S P R true 8 10, I 12 P R true 3, 7, 11, E Sequent calculus, with ATP s E Let the following derivation be called D: 26 October 2011 6 of 18
Γ h 1 :P Q Γ, u:q h 1 :P Q WEAK Γ, u:q u:q VAR proof 1.1 Γ, u:q λx. u(h 1 x) : P Γ, u:q inl(λx. u(h 1 x)) :(P ) (R ) I L And let the following derivation be called E : Γ h 2 :R S Γ, v:s h 2 :R S WEAK Γ, v:s v:s VAR proof 1.1 Γ, v:s λy. v(h 2 y) : R Γ, v:s inr(λy. v(h 2 y)) :(P ) (R ) I R And by disjunction elimination we get the conclusion: Γ h 3 :(Q ) (S ) Γ h 1 :P Q D Γ, u:q... Γ h 2 :R S E Γ, v:s... Γ case h 3 of inl u inl(λx. u(h 1 x)) inr v inr(λy. v(h 2 y)) :(P ) (R ) Eu,v However, this last step is too big to write out in full and fit on the page. 1.6 Strengthening the Antecedent (SA): P Q P R Q Naturally, we ll also have a proof of P Q R P Q. Rather than using commutativity of conjunction followed by SA, we could prove it directly using the same idea as we have here. Natural deduction P Qtrue P R true u P true Q true P R Qtrue E L I u 26 October 2011 7 of 18
Fitch-style 1 P Qtrue Premise 2 P R true Assumption 3 P true 2, E L 4 Q true 1, 3, 5 P R Qtrue 2 4, I Sequent calculus N.B., this is just the hypothetical syllogism with fst for h 1 (and h 1 for h 2 ). VAR Γ h 1 :P Q Γ, u:p R u:p R Γ, u:p R h 1 :P Q WEAK Γ, u:p R fst u : P E L Γ, u:p R h 1 (fst u) : Q Γ λu. h 1 (fst u) : P R Q I 1.7 Weakening the Consequent (WC): P Q P Q R Naturally, we ll also have a proof of P Q P R Q. Rather than using WC followed by commutativity of disjunction, we could prove it directly using the same idea as we have here. Natural deduction P Qtrue P true u Q true Q R true P Q R true I L I u Fitch-style 1 P Qtrue Premise 2 P true Assumption 3 Q true 1, 2, 4 Q R true 3, I L 5 P Q R true 2 4, I 26 October 2011 8 of 18
Sequent calculus N.B., this is just the hypothetical syllogism with inl for h 2. Γ h 1 :P Q Γ, u:p h 1 :P Q WEAK Γ, u:p u:p VAR Γ, u:p h 1 u : Q Γ, u:p inl(h 1 u) : Q R Γ λu. inl(h 1 u) : P Q R I L I 2 Okay, by now it should be clear how to convert between the three styles of proof. From now on I ll only give one style. 2.1 A C B G,A (E D) A B 1 A C B G true Premise 2 A (E D) true Premise 3 A true 2, E L 4 A C true 3, I L 5 B G true 1, 4, 6 B true 5, E L 7 A B true 3, 6, I 2.2 (P R) M P,P M P R,P M R P 1 (P R) M P true Premise 2 P M P Rtrue Premise 3 P M true Premise 4 P Rtrue 2, 3, 5 M P true 1, 4, 6 R P true 3, 4, 5, CD (proof 1.4) 26 October 2011 9 of 18
2.3 F C B D, B A F, B E C, B D 1 F C B D true Premise 2 B A F true Premise 3 B E C true Premise 4 B true Premise 5 B A true 4, I L 6 F true 2, 5, 7 B E true 4, I L 8 C true 3, 7, 9 F C true 6, 8, I 10 B D true 1, 9, 11 D true 4, 10, DS (proof 1.2) 2.4 A B,A C D,D F F 1 A B true Premise 2 A C Dtrue Premise 3 D F true Premise 4 A true 1, E L 5 A C true 4, I L 6 D true 2, 5, 7 F true 3, 6, 26 October 2011 10 of 18
2.5 A B,C D,A C F E G 1 A B true Premise 2 C D true Premise 3 A C F true Premise 4 A true 1, E L 5 C true 2, E L 6 A C true 4, 5, I 7 E F true 3, 6, 8 E true 7, E L 9 E G true 8, I L 2.6 A B C D,C E, E,C D A 1 A B C D true Premise 2 C E true Premise 3 E true Premise 4 C Dtrue Premise 5 C true 2, 3, DS (proof 1.2) 6 D true 4, 5, 7 C D true 5, 6, I 8 (A B C D) (C D A B) true 1, Notation 9 C D A B true 8, E R 10 A B true 7, 9, 11 A true 10, E L 26 October 2011 11 of 18
2.7 A B, B,A C D,C E F, F E 1 A B true Premise 2 B true Premise 3 A C D true Premise 4 C E F true Premise 5 F true Premise 6 A true 1, 2, DS (proof 1.2) 7 C D true 3, 6, 8 C true 7, E L 9 (C F) (E F C) true 4, Notation 10 C F true 9, E L 11 E F true 8, 10, 12 E true 5, 11, DS (proof 1.2) 2.8 A C, C, A C D D 1 A C true Premise 2 C true Premise 3 A C D true Premise 4 A true 1, 2, MT (proof 1.1) 5 C D true 3, 4, 6 D true 2, 5, DS (proof 1.2) 26 October 2011 12 of 18
2.9 A B C,C, A B 1 A B C true Premise 2 C true Premise 3 A true Premise 4 A B true 3, I L 5 C true 1, 4, 6 true 2, 5, E 7 B true 6, E 2.10 A B,B, A D E F, F E While we don t have double negation elimination in intuitionistic logic, we do have double negation introduction. Lemma 2.10.1 (Double Negation Introduction). For all φ we have that φ φ. Proof. 1 φ true Premise 2 φ true Assumption 3 true 1, 2, E 4 φ true 2 3, I Using double negation introduction we can prove a variation on modus tollens with the polarity reversed on the one premise and on the other premise s consequent. Lemma 2.10.2. For all φ and ψ we have that φ ψ,ψ φ. Proof. 1 φ ψ true Premise 2 ψ true Premise 3 ψ true 2, Lemma 2.10.1 4 φ true 1, 3, MT (proof 1.1) 26 October 2011 13 of 18
1 A B true Premise 2 B true Premise 3 A D E F true Premise 4 F true Premise 5 A true 1, 2, Lemma 2.10.2 6 A D true 5, I L 7 ( A D F) (E F A D) true 3, Notation 8 A D F true 7, E L 9 E F true 6, 8, 10 E true 4, 9, DS (proof 1.2) 2.11 A B C, C D,A D A 1 A B C true Premise 2 C D true Premise 3 A D true Premise 4 C true Assumption 5 (A B) true 1, 4, MT (proof 1.1) 6 A B true 5, De Morgan s law (see handout) 7 A true 6, E L 8 C A true 4 7, I 9 D true Assumption 10 A true 3, 9, Lemma 2.10.2 11 D A true 9 10, I 12 A true 2, 8, 11, E 2.12 B C,A C A B D This cannot be proven without using classical axioms. (Proving that to be the case is left as an exercise for the reader. Though you may want to check out http://hackage.haskell.org/ package/djinn along the way.) 26 October 2011 14 of 18
1 B C true Premise 2 A C true Premise 3 A true Assumption 4 C true 2, 3, 5 B true 1, 4, Lemma 2.10.2 6 B true 5, C 7 B D true 6, I L 8 A B D true 3 7, I An alternative proof is to use negation introduction instead of Lemma 2.10.2 to derive B true. 1 B C true Premise 2 A C true Premise 3 A true Assumption 4 C true 2, 3, 5 B true Assumption 6 C true 1, 5, 7 true 4, 6, E 8 B true 5 7, I 9 B true 8, C 10 B D true 9, I L 11 A B D true 3 10, I 26 October 2011 15 of 18
2.13 B E D, E B D F 1 B E D true Premise 2 E true Premise 3 B true Assumption 4 (B D) (E D B) true 1, Notation 5 B D true 4, E L 6 E D true 3, 5, 7 D true 2, 6, DS (proof 1.2) 8 D F true 7, I L 9 B D F true 3 8, I 2.14 A B,A C,B D,D C E 1 A B true Premise 2 A C true Premise 3 B Dtrue Premise 4 D true Premise 5 (B D) (D B) true 3, Notation 6 B Dtrue 5, E L 7 B true 4, 6, HS (proof 1.3) 8 C E true 1, 2, 7, CD (proof 1.4) 26 October 2011 16 of 18
2.15 A B,C B,A C A 1 A Btrue Premise 2 C B true Premise 3 A C true Premise 4 A true Assumption 5 C true Assumption 6 C true 3, 4, 7 true 5, 6, E 8 C true 5 7, I 9 B true Assumption 10 A true 1, 9, MT (proof 1.1) 11 true 4, 10, E 12 B true 9 11, I 13 true 2, 8, 12, E 14 A true 4 13, I 3 Let, SE=Superman exists EE=evil exists / is not prevented SWPE=Superman is willing to prevent evil SAPE=Superman is able to prevent evil SI = Superman is impotent SM = Superman is malevolent 26 October 2011 17 of 18
Thus, with each sentence given in order as a premise, one argument is as follows. 1 SWPE SAPE EE true Premise 2 SAPE SI true Premise 3 SWPE SM true Premise 4 EE true Premise 5 SE SI SM true Premise 6 (SWPE SAPE) true 1, 4, Lemma 2.10.2 7 SE true Assumption 8 SI SM true 5, 7, 9 SI true 8, E L 10 SAPE true 2, 9, MT (proof 1.1) 11 SAPE true 10, C 12 SM true 8, E R 13 SWPE true 3, 12, MT (proof 1.1) 14 SWPE true 13, C 15 SWPE SAPE true 11, 14, I 16 true 6, 15, E 17 SE true 7 16, I I don t think there s an intuitionistic proof of this, though I haven t proven there isn t. We could avoid the uses of C if we could prove either φ ψ, ψ φ (instead of using modus tollens and double-negation elimination) or (φ ψ) φ ψ (and then use the negations of both disjuncts to derive the contradiction); however, both of those theorems can only be proven in a classical setting. 26 October 2011 18 of 18