Q250, Homework 7: 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 (W EAK) 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 (h1 , h2 , . . . ) 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 = ·. def
And finally, for sequent calculus I’m just going to assume the definition ¬φ = φ ⇒ ⊥, and so I’ll be using ⇒I and ⇒E instead of ¬I and ¬E.
1 Alternative rules 1.1 Modus Tollens (MT): P ⇒ Q, ¬Q ⊢ ¬P Natural deduction P ⇒ Q true Q true
u
P true ⇒E ⊥ true ¬P true
26 October 2011
¬Q true ¬E ¬I u
1 of 18
Q250, Homework 7 Fitch-style 1
P ⇒ Q true
Premise
2
¬Q true
Premise
3
P true
Assumption
4
Q true
1, 3, ⇒E
5
⊥ true
2, 4, ¬E
¬P true
6
3–5, ¬I
Sequent calculus N.B., this is the same as the hypothetical syllogism with R = ⊥. Γ ⊢ h1 : P ⇒ Q W EAK VAR Γ, u : P ⊢ h1 : P ⇒ Q Γ, u : P ⊢ u : P Γ ⊢ h2 : Q ⇒ ⊥ ⇒E W EAK Γ, u : P ⊢ h1 u : Q Γ, u : P ⊢ h2 : Q ⇒ ⊥ ⇒E Γ, u : P ⊢ h2 (h1 u) : ⊥ ⇒I Γ ⊢ λu. h2 (h1 u) : P ⇒ ⊥
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 true P ∨ Q true
u
¬P true ¬E
⊥ true ⊥E Q true ⇒I u P ⇒ Q true Q true
v
Q true ⇒I v Q ⇒ Q true
∨E
Natural deduction with ATP’s ∨E and ¬E P ∨ Q true
26 October 2011
P true
u
¬P true
Q true Q true
¬E
Q true
v ∨E u,v
2 of 18
Q250, Homework 7 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 ⇒ Q true
7
Q true
Assumption
8
Q true
7
9
Q ⇒ Q true
7–8, ⇒I
Q true
1, 6, 9, ∨E
10
3–5, ⇒I
Sequent calculus, with the implicational ∨E Γ ⊢ h2 : P ⇒ ⊥ W EAK Γ, u : P ⊢ u : P Γ, u : P ⊢ h2 : P ⇒ ⊥ ⇒E Γ, u : P ⊢ h2 u : ⊥ ⊥E Γ, u : P ⊢ abort (h2 u) : Q ⇒I Γ ⊢ λu. abort (h2 u) : P ⇒ Q Γ ⊢ either (λu. abort (h2 u)) (λv. v) h1 : Q VAR
Γ ⊢ h1 : P ∨ Q
VAR
Γ, v : Q ⊢ v : Q ⇒I Γ ⊢ λv. v : Q ⇒ Q
Sequent calculus, with ATP’s ∨E Γ ⊢ h2 : P ⇒ ⊥ W EAK Γ, u : P ⊢ u : P Γ, u : P ⊢ h2 : P ⇒ ⊥ ⇒E Γ, u : P ⊢ h2 u : ⊥ ⊥E Γ, u : P ⊢ abort (h2 u) : Q Γ ⊢ case h1 of inl u ⇒ abort (h2 u) | inr v ⇒ v : Q VAR
Γ ⊢ h1 : P ∨ 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 ⇒ Q true Q true
⇒E
R true P ⇒ R true
26 October 2011
Q ⇒ R true ⇒E ⇒I u
3 of 18
∨E
Q250, Homework 7 Fitch-style 1
P ⇒ Q true
Premise
2
Q ⇒ R true
Premise
3
P true
Assumption
4
Q true
1, 3, ⇒E
5
R true
2, 4, ⇒E
6
P ⇒ R true
3–5, ⇒I
Sequent calculus N.B., this is just function composition. Γ ⊢ h1 : P ⇒ Q W EAK Γ, u : P ⊢ u : P Γ, u : P ⊢ h1 : P ⇒ Q Γ ⊢ h2 : Q ⇒ R ⇒E W EAK Γ, u : P ⊢ h1 u : Q Γ, u : P ⊢ h2 : Q ⇒ R ⇒E Γ, u : P ⊢ h2 (h1 u) : R ⇒I Γ ⊢ λu. h2 (h1 u) : P ⇒ R VAR
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 ⇒ Q true P ∨ R true
R ⇒ S true
u
P true ⇒E
Q true Q ∨ S true
∨IL
Q ∨ S true
26 October 2011
v
R true ⇒E
S true Q ∨ S true
∨IR ∨E u,v
4 of 18
Q250, Homework 7 Fitch-style 1
P ⇒ Q true
Premise
2
R ⇒ S true
Premise
3
P ∨ R true
Premise
4
P true
Assumption
5
Q true
1, 4, ⇒E
6
Q ∨ S true
5, ∨IL
7
P ⇒ Q ∨ S true
4–6, ⇒I
8
R true
Assumption
9
S true
2, 8, ⇒E
10
Q ∨ S true
9, ∨IR
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 ⊢ h1 : P ⇒ Q W EAK VAR Γ, u : P ⊢ h1 : P ⇒ Q Γ, u : P ⊢ u : P ⇒E Γ, u : P ⊢ h1 u : Q ∨IL Γ, u : P ⊢ inl (h1 u) : Q ∨ S And let the following derivation be called E : Γ ⊢ h2 : R ⇒ S W EAK VAR Γ, v : R ⊢ h2 : R ⇒ S Γ, v : R ⊢ v : R ⇒E Γ, v : P ⊢ h2 v : S ∨IR Γ, v : P ⊢ inr (h2 v) : Q ∨ S And finally we put those together with ATP-style disjunction elimination.
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Q250, Homework 7
Γ ⊢ h3 : P ∨ R
Γ ⊢ h1 : P ⇒ Q
Γ ⊢ h2 : R ⇒ S
D
E
Γ, u : P ⊢ inl (h1 u) : Q ∨ S
Γ, v : P ⊢ inr (h2 v) : Q ∨ S
Γ ⊢ case h3 of inl u ⇒ inl (h1 u) | inr v ⇒ inr (h2 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 P ⇒ Q true
u
¬Q true proof 1.1 ¬P true ∨IL ¬P ∨ ¬R true ¬P ∨ ¬R true
¬Q ∨ ¬S true
R ⇒ S true
v
¬S true proof 1.1 ¬R true ∨IR ¬P ∨ ¬R true ∨E u,v
Fitch-style 1
P ⇒ Q true
Premise
2
R ⇒ S true
Premise
3
¬Q ∨ ¬S true
Premise
4
¬Q true
Assumption
5
¬P true
1, 4, MT (proof 1.1)
6
¬P ∨ ¬R true
5, ∨IL
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, ∨IR
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
Q250, Homework 7
Γ ⊢ h1 : P ⇒ Q W EAK VAR Γ, u : Q ⇒ ⊥ ⊢ h1 : P ⇒ Q Γ, u : Q ⇒ ⊥ ⊢ u : Q ⇒ ⊥ proof 1.1 Γ, u : Q ⇒ ⊥ ⊢ λx. u (h1 x) : P ⇒ ⊥ ∨IL Γ, u : Q ⇒ ⊥ ⊢ inl (λx. u (h1 x)) : (P ⇒ ⊥) ∨ (R ⇒ ⊥) And let the following derivation be called E : Γ ⊢ h2 : R ⇒ S W EAK VAR Γ, v : S ⇒ ⊥ ⊢ h2 : R ⇒ S Γ, v : S ⊢ v : S ⇒ ⊥ proof 1.1 Γ, v : S ⇒ ⊥ ⊢ λy. v (h2 y) : R ⇒ ⊥ ∨IR Γ, v : S ⇒ ⊥ ⊢ inr (λy. v (h2 y)) : (P ⇒ ⊥) ∨ (R ⇒ ⊥) And by disjunction elimination we get the conclusion:
Γ ⊢ h3 : (Q ⇒ ⊥) ∨ (S ⇒ ⊥)
Γ ⊢ h1 : P ⇒ Q
Γ ⊢ h2 : R ⇒ S
D
E
Γ, u : Q ⇒ ⊥ ⊢ . . .
Γ, v : S ⇒ ⊥ ⊢ . . .
Γ ⊢ case h3 of inl u ⇒ inl (λx. u (h1 x)) | inr v ⇒ inr (λy. v (h2 y)) : (P ⇒ ⊥) ∨ (R ⇒ ⊥)
∨E u,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 u
P ∧ R true ∧E P ⇒ Q true L P true ⇒E Q true ⇒I u P ∧ R ⇒ Q true
26 October 2011
7 of 18
Q250, Homework 7 Fitch-style 1
P ⇒ Q true
Premise
2
P ∧ R true
Assumption
3
P true
2, ∧EL
4
Q true
1, 3, ⇒E
5
P ∧ R ⇒ Q true
2–4, ⇒I
Sequent calculus N.B., this is just the hypothetical syllogism with fst for h1 (and h1 for h2 ).
VAR
Γ, u : P ∧ R ⊢ u : P ∧ R Γ ⊢ h1 : P ⇒ Q W EAK ∧EL Γ, u : P ∧ R ⊢ h1 : P ⇒ Q Γ, u : P ∧ R ⊢ fst u : P ⇒E Γ, u : P ∧ R ⊢ h1 (fst u) : Q ⇒I Γ ⊢ λu. h1 (fst u) : P ∧ R ⇒ Q
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 ⇒ Q true
u
P true ⇒E
Q true Q ∨ R true P ⇒ Q ∨ R true
∨IL ⇒I u
Fitch-style
26 October 2011
1
P ⇒ Q true
2
P true
Assumption
3
Q true
1, 2, ⇒E
4
Q ∨ R true
3, ∨IL
5
P ⇒ Q ∨ R true
Premise
2–4, ⇒I
8 of 18
Q250, Homework 7 Sequent calculus N.B., this is just the hypothetical syllogism with inl for h2 . Γ ⊢ h1 : P ⇒ Q W EAK VAR Γ, u : P ⊢ h1 : P ⇒ Q Γ, u : P ⊢ u : P ⇒E Γ, u : P ⊢ h1 u : Q ∨IL Γ, u : P ⊢ inl (h1 u) : Q ∨ R ⇒I Γ ⊢ λu. inl (h1 u) : P ⇒ Q ∨ R
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, ∧EL
4
A ∨C true
3, ∨IL
5
B ∧ G true
1, 4, ⇒E
6
B true
5, ∧EL
7
A ∧ B true
3, 6, ∧I
2.2 (P ⇒ R) ⇒ M ⇒ P, P ∨ M ⇒ P ⇒ R, P ∨ M ⊢ R ∨ P
26 October 2011
1
(P ⇒ R) ⇒ M ⇒ P true
Premise
2
P ∨ M ⇒ P ⇒ R true
Premise
3
P ∨ M true
Premise
4
P ⇒ R true
2, 3, ⇒E
5
M ⇒ P true
1, 4, ⇒E
6
R ∨ P true
3, 4, 5, CD (proof 1.4)
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Q250, Homework 7
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, ∨IL
6
F true
2, 5, ⇒E
7
¬B ∨ E true
4, ∨IL
8
C true
3, 7, ⇒E
9
F ∧C true
6, 8, ∧I
10
B ∨ D true
1, 9, ⇒E
11
D true
4, 10, DS (proof 1.2)
2.4 A ∧ B, A ∨C ⇒ D, D ⇒ F ⊢ F
26 October 2011
1
A ∧ B true
Premise
2
A ∨C ⇒ D true
Premise
3
D ⇒ F true
Premise
4
A true
1, ∧EL
5
A ∨C true
4, ∨IL
6
D true
2, 5, ⇒E
7
F true
3, 6, ⇒E
10 of 18
Q250, Homework 7
2.5 A ∧ B,C ∧ D, A ∧C ⇒ E ∧ F ⊢ E ∨ G 1
A ∧ B true
Premise
2
C ∧ D true
Premise
3
A ∧C ⇒ E ∧ F true
Premise
4
A true
1, ∧EL
5
C true
2, ∧EL
6
A ∧C true
4, 5, ∧I
7
E ∧ F true
3, 6, ⇒E
8
E true
7, ∧EL
9
E ∨ G true
8, ∨IL
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 ⇒ D true
Premise
5
C true
2, 3, DS (proof 1.2)
6
D true
4, 5, ⇒E
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, ∧ER
10
A ∧ B true
7, 9, ⇒E
11
A true
10, ∧EL
26 October 2011
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Q250, Homework 7
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, ⇒E
8
C true
7, ∧EL
9
(C ⇒ E ∨ F) ∧ (E ∨ F ⇒ C) true
4, Notation
10
C ⇒ E ∨ F true
9, ∧EL
11
E ∨ F true
8, 10, ⇒E
12
E true
5, 11, DS (proof 1.2)
2.8 A ⇒ C, ¬C, ¬A ⇒ C ∨ D ⊢ D
26 October 2011
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, ⇒E
6
D true
2, 5, DS (proof 1.2)
12 of 18
Q250, Homework 7
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, ∨IL
5
¬C true
1, 4, ⇒E
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
Q250, Homework 7
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, ∨IL
7
(¬A ∨ D ⇒ E ∨ F) ∧ (E ∨ F ⇒ ¬A ∨ D) true
3, Notation
8
¬A ∨ D ⇒ E ∨ F true
7, ∧EL
9
E ∨ F true
6, 8, ⇒E
E true
4, 9, DS (proof 1.2)
10
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, ∧EL
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.)
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Q250, Homework 7
1
¬B ⇒ ¬C true
Premise
2
A ⇒ C true
Premise
3
A true
Assumption
4
C true
2, 3, ⇒E
5
¬¬B true
1, 4, Lemma 2.10.2
6
B true
5, ¬¬C
7
B ∨ D true
6, ∨IL
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.
26 October 2011
1
¬B ⇒ ¬C true
Premise
2
A ⇒ C true
Premise
3
A true
Assumption
4
C true
2, 3, ⇒E
5
¬B true
Assumption
6
¬C true
1, 5, ⇒E
7
⊥ true
4, 6, ¬E
8
¬¬B true
5–7, ¬I
9
B true
8, ¬¬C
10
B ∨ D true
9, ∨IL
11
A ⇒ B ∨ D true
3–10, ⇒I
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Q250, Homework 7
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 ⇒ E ∨ D) ∧ (E ∨ D ⇒ B) true
1, Notation
5
B ⇒ E ∨ D true
4, ∧EL
6
E ∨ D true
3, 5, ⇒E
7
D true
2, 6, DS (proof 1.2)
8
D ∨ F true
7, ∨IL
9
B ⇒ D ∨ F true
3–8, ⇒I
2.14 A ∨ B, A ⇒ C, B ⇔ D, D ⇒ E ⊢ C ∨ E
26 October 2011
1
A ∨ B true
Premise
2
A ⇒ C true
Premise
3
B ⇔ D true
Premise
4
D ⇒ E true
Premise
5
(B ⇒ D) ∧ (D ⇒ B) true
3, Notation
6
B ⇒ D true
5, ∧EL
7
B ⇒ E true
4, 6, HS (proof 1.3)
8
C ∨ E true
1, 2, 7, CD (proof 1.4)
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Q250, Homework 7
2.15 A ⇒ B,C ∨ ¬B, A ⇒ ¬C ⊢ ¬A 1
A ⇒ B true
Premise
2
C ∨ ¬B true
Premise
3
A ⇒ ¬C true
Premise
4
A true
Assumption
5
C true
Assumption
6
¬C true
3, 4, ⇒E
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
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17 of 18
Q250, Homework 7 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, ⇒E
9
¬SI true
8, ∧EL
10
¬¬SAPE true
2, 9, MT (proof 1.1)
11
SAPE true
10, ¬¬C
12
¬SM true
8, ∧ER
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
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