(* Title: HOL/ex/Intuitionistic.thy ID: $Id: Intuitionistic.thy,v 1.4 2005/09/22 21:56:47 nipkow Exp $ Author: Lawrence C Paulson, Cambridge University Computer Laboratory Copyright 1991 University of Cambridge Taken from FOL/ex/int.ML *) header {* Higher-Order Logic: Intuitionistic predicate calculus problems *} theory Intuitionistic imports Main begin (*Metatheorem (for PROPOSITIONAL formulae...): P is classically provable iff ~~P is intuitionistically provable. Therefore ~P is classically provable iff it is intuitionistically provable. Proof: Let Q be the conjuction of the propositions A|~A, one for each atom A in P. Now ~~Q is intuitionistically provable because ~~(A|~A) is and because ~~ distributes over &. If P is provable classically, then clearly Q-->P is provable intuitionistically, so ~~(Q-->P) is also provable intuitionistically. The latter is intuitionistically equivalent to ~~Q-->~~P, hence to ~~P, since ~~Q is intuitionistically provable. Finally, if P is a negation then ~~P is intuitionstically equivalent to P. [Andy Pitts] *) lemma "(~~(P&Q)) = ((~~P) & (~~Q))" by iprover lemma "~~ ((~P --> Q) --> (~P --> ~Q) --> P)" by iprover (* ~~ does NOT distribute over | *) lemma "(~~(P-->Q)) = (~~P --> ~~Q)" by iprover lemma "(~~~P) = (~P)" by iprover lemma "~~((P --> Q | R) --> (P-->Q) | (P-->R))" by iprover lemma "(P=Q) = (Q=P)" by iprover lemma "((P --> (Q | (Q-->R))) --> R) --> R" by iprover lemma "(((G-->A) --> J) --> D --> E) --> (((H-->B)-->I)-->C-->J) --> (A-->H) --> F --> G --> (((C-->B)-->I)-->D)-->(A-->C) --> (((F-->A)-->B) --> I) --> E" by iprover (* Lemmas for the propositional double-negation translation *) lemma "P --> ~~P" by iprover lemma "~~(~~P --> P)" by iprover lemma "~~P & ~~(P --> Q) --> ~~Q" by iprover (* de Bruijn formulae *) (*de Bruijn formula with three predicates*) lemma "((P=Q) --> P&Q&R) & ((Q=R) --> P&Q&R) & ((R=P) --> P&Q&R) --> P&Q&R" by iprover (*de Bruijn formula with five predicates*) lemma "((P=Q) --> P&Q&R&S&T) & ((Q=R) --> P&Q&R&S&T) & ((R=S) --> P&Q&R&S&T) & ((S=T) --> P&Q&R&S&T) & ((T=P) --> P&Q&R&S&T) --> P&Q&R&S&T" by iprover (*** Problems from Sahlin, Franzen and Haridi, An Intuitionistic Predicate Logic Theorem Prover. J. Logic and Comp. 2 (5), October 1992, 619-656. ***) (*Problem 1.1*) lemma "(ALL x. EX y. ALL z. p(x) & q(y) & r(z)) = (ALL z. EX y. ALL x. p(x) & q(y) & r(z))" by (iprover del: allE elim 2: allE') (*Problem 3.1*) lemma "~ (EX x. ALL y. p y x = (~ p x x))" by iprover (* Intuitionistic FOL: propositional problems based on Pelletier. *) (* Problem ~~1 *) lemma "~~((P-->Q) = (~Q --> ~P))" by iprover (* Problem ~~2 *) lemma "~~(~~P = P)" by iprover (* Problem 3 *) lemma "~(P-->Q) --> (Q-->P)" by iprover (* Problem ~~4 *) lemma "~~((~P-->Q) = (~Q --> P))" by iprover (* Problem ~~5 *) lemma "~~((P|Q-->P|R) --> P|(Q-->R))" by iprover (* Problem ~~6 *) lemma "~~(P | ~P)" by iprover (* Problem ~~7 *) lemma "~~(P | ~~~P)" by iprover (* Problem ~~8. Peirce's law *) lemma "~~(((P-->Q) --> P) --> P)" by iprover (* Problem 9 *) lemma "((P|Q) & (~P|Q) & (P| ~Q)) --> ~ (~P | ~Q)" by iprover (* Problem 10 *) lemma "(Q-->R) --> (R-->P&Q) --> (P-->(Q|R)) --> (P=Q)" by iprover (* 11. Proved in each direction (incorrectly, says Pelletier!!) *) lemma "P=P" by iprover (* Problem ~~12. Dijkstra's law *) lemma "~~(((P = Q) = R) = (P = (Q = R)))" by iprover lemma "((P = Q) = R) --> ~~(P = (Q = R))" by iprover (* Problem 13. Distributive law *) lemma "(P | (Q & R)) = ((P | Q) & (P | R))" by iprover (* Problem ~~14 *) lemma "~~((P = Q) = ((Q | ~P) & (~Q|P)))" by iprover (* Problem ~~15 *) lemma "~~((P --> Q) = (~P | Q))" by iprover (* Problem ~~16 *) lemma "~~((P-->Q) | (Q-->P))" by iprover (* Problem ~~17 *) lemma "~~(((P & (Q-->R))-->S) = ((~P | Q | S) & (~P | ~R | S)))" oops (*Dijkstra's "Golden Rule"*) lemma "(P&Q) = (P = (Q = (P|Q)))" by iprover (****Examples with quantifiers****) (* The converse is classical in the following implications... *) lemma "(EX x. P(x)-->Q) --> (ALL x. P(x)) --> Q" by iprover lemma "((ALL x. P(x))-->Q) --> ~ (ALL x. P(x) & ~Q)" by iprover lemma "((ALL x. ~P(x))-->Q) --> ~ (ALL x. ~ (P(x)|Q))" by iprover lemma "(ALL x. P(x)) | Q --> (ALL x. P(x) | Q)" by iprover lemma "(EX x. P --> Q(x)) --> (P --> (EX x. Q(x)))" by iprover (* Hard examples with quantifiers *) (*The ones that have not been proved are not known to be valid! Some will require quantifier duplication -- not currently available*) (* Problem ~~19 *) lemma "~~(EX x. ALL y z. (P(y)-->Q(z)) --> (P(x)-->Q(x)))" by iprover (* Problem 20 *) lemma "(ALL x y. EX z. ALL w. (P(x)&Q(y)-->R(z)&S(w))) --> (EX x y. P(x) & Q(y)) --> (EX z. R(z))" by iprover (* Problem 21 *) lemma "(EX x. P-->Q(x)) & (EX x. Q(x)-->P) --> ~~(EX x. P=Q(x))" by iprover (* Problem 22 *) lemma "(ALL x. P = Q(x)) --> (P = (ALL x. Q(x)))" by iprover (* Problem ~~23 *) lemma "~~ ((ALL x. P | Q(x)) = (P | (ALL x. Q(x))))" by iprover (* Problem 25 *) lemma "(EX x. P(x)) & (ALL x. L(x) --> ~ (M(x) & R(x))) & (ALL x. P(x) --> (M(x) & L(x))) & ((ALL x. P(x)-->Q(x)) | (EX x. P(x)&R(x))) --> (EX x. Q(x)&P(x))" by iprover (* Problem 27 *) lemma "(EX x. P(x) & ~Q(x)) & (ALL x. P(x) --> R(x)) & (ALL x. M(x) & L(x) --> P(x)) & ((EX x. R(x) & ~ Q(x)) --> (ALL x. L(x) --> ~ R(x))) --> (ALL x. M(x) --> ~L(x))" by iprover (* Problem ~~28. AMENDED *) lemma "(ALL x. P(x) --> (ALL x. Q(x))) & (~~(ALL x. Q(x)|R(x)) --> (EX x. Q(x)&S(x))) & (~~(EX x. S(x)) --> (ALL x. L(x) --> M(x))) --> (ALL x. P(x) & L(x) --> M(x))" by iprover (* Problem 29. Essentially the same as Principia Mathematica *11.71 *) lemma "(((EX x. P(x)) & (EX y. Q(y))) --> (((ALL x. (P(x) --> R(x))) & (ALL y. (Q(y) --> S(y)))) = (ALL x y. ((P(x) & Q(y)) --> (R(x) & S(y))))))" by iprover (* Problem ~~30 *) lemma "(ALL x. (P(x) | Q(x)) --> ~ R(x)) & (ALL x. (Q(x) --> ~ S(x)) --> P(x) & R(x)) --> (ALL x. ~~S(x))" by iprover (* Problem 31 *) lemma "~(EX x. P(x) & (Q(x) | R(x))) & (EX x. L(x) & P(x)) & (ALL x. ~ R(x) --> M(x)) --> (EX x. L(x) & M(x))" by iprover (* Problem 32 *) lemma "(ALL x. P(x) & (Q(x)|R(x))-->S(x)) & (ALL x. S(x) & R(x) --> L(x)) & (ALL x. M(x) --> R(x)) --> (ALL x. P(x) & M(x) --> L(x))" by iprover (* Problem ~~33 *) lemma "(ALL x. ~~(P(a) & (P(x)-->P(b))-->P(c))) = (ALL x. ~~((~P(a) | P(x) | P(c)) & (~P(a) | ~P(b) | P(c))))" oops (* Problem 36 *) lemma "(ALL x. EX y. J x y) & (ALL x. EX y. G x y) & (ALL x y. J x y | G x y --> (ALL z. J y z | G y z --> H x z)) --> (ALL x. EX y. H x y)" by iprover (* Problem 39 *) lemma "~ (EX x. ALL y. F y x = (~F y y))" by iprover (* Problem 40. AMENDED *) lemma "(EX y. ALL x. F x y = F x x) --> ~(ALL x. EX y. ALL z. F z y = (~ F z x))" by iprover (* Problem 44 *) lemma "(ALL x. f(x) --> (EX y. g(y) & h x y & (EX y. g(y) & ~ h x y))) & (EX x. j(x) & (ALL y. g(y) --> h x y)) --> (EX x. j(x) & ~f(x))" by iprover (* Problem 48 *) lemma "(a=b | c=d) & (a=c | b=d) --> a=d | b=c" by iprover (* Problem 51 *) lemma "((EX z w. (ALL x y. (P x y = ((x = z) & (y = w))))) --> (EX z. (ALL x. (EX w. ((ALL y. (P x y = (y = w))) = (x = z))))))" by iprover (* Problem 52 *) (*Almost the same as 51. *) lemma "((EX z w. (ALL x y. (P x y = ((x = z) & (y = w))))) --> (EX w. (ALL y. (EX z. ((ALL x. (P x y = (x = z))) = (y = w))))))" by iprover (* Problem 56 *) lemma "(ALL x. (EX y. P(y) & x=f(y)) --> P(x)) = (ALL x. P(x) --> P(f(x)))" by iprover (* Problem 57 *) lemma "P (f a b) (f b c) & P (f b c) (f a c) & (ALL x y z. P x y & P y z --> P x z) --> P (f a b) (f a c)" by iprover (* Problem 60 *) lemma "ALL x. P x (f x) = (EX y. (ALL z. P z y --> P z (f x)) & P x y)" by iprover end
lemma
(¬ ¬ (P ∧ Q)) = (¬ ¬ P ∧ ¬ ¬ Q)
lemma
¬ ¬ ((¬ P --> Q) --> (¬ P --> ¬ Q) --> P)
lemma
(¬ ¬ (P --> Q)) = (¬ ¬ P --> ¬ ¬ Q)
lemma
(¬ ¬ ¬ P) = (¬ P)
lemma
¬ ¬ ((P --> Q ∨ R) --> (P --> Q) ∨ (P --> R))
lemma
(P = Q) = (Q = P)
lemma
((P --> Q ∨ (Q --> R)) --> R) --> R
lemma
(((G --> A) --> J) --> D --> E) -->
(((H --> B) --> I) --> C --> J) -->
(A --> H) -->
F -->
G -->
(((C --> B) --> I) --> D) --> (A --> C) --> (((F --> A) --> B) --> I) --> E
lemma
P --> ¬ ¬ P
lemma
¬ ¬ (¬ ¬ P --> P)
lemma
¬ ¬ P ∧ ¬ ¬ (P --> Q) --> ¬ ¬ Q
lemma
(P = Q --> P ∧ Q ∧ R) ∧ (Q = R --> P ∧ Q ∧ R) ∧ (R = P --> P ∧ Q ∧ R) -->
P ∧ Q ∧ R
lemma
(P = Q --> P ∧ Q ∧ R ∧ S ∧ T) ∧
(Q = R --> P ∧ Q ∧ R ∧ S ∧ T) ∧
(R = S --> P ∧ Q ∧ R ∧ S ∧ T) ∧
(S = T --> P ∧ Q ∧ R ∧ S ∧ T) ∧ (T = P --> P ∧ Q ∧ R ∧ S ∧ T) -->
P ∧ Q ∧ R ∧ S ∧ T
lemma
(∀x. ∃y. ∀z. p x ∧ q y ∧ r z) = (∀z. ∃y. ∀x. p x ∧ q y ∧ r z)
lemma
¬ (∃x. ∀y. p y x = (¬ p x x))
lemma
¬ (P --> Q) ≠ (¬ Q --> ¬ P)
lemma
¬ ¬ ¬ P ≠ P
lemma
¬ (P --> Q) --> Q --> P
lemma
¬ (¬ P --> Q) ≠ (¬ Q --> P)
lemma
¬ ¬ ((P ∨ Q --> P ∨ R) --> P ∨ (Q --> R))
lemma
¬ ¬ (P ∨ ¬ P)
lemma
¬ ¬ (P ∨ ¬ ¬ ¬ P)
lemma
¬ ¬ (((P --> Q) --> P) --> P)
lemma
(P ∨ Q) ∧ (¬ P ∨ Q) ∧ (P ∨ ¬ Q) --> ¬ (¬ P ∨ ¬ Q)
lemma
(Q --> R) --> (R --> P ∧ Q) --> (P --> Q ∨ R) --> P = Q
lemma
P = P
lemma
¬ ((P = Q) = R) ≠ (P = (Q = R))
lemma
(P = Q) = R --> ¬ P ≠ (Q = R)
lemma
(P ∨ Q ∧ R) = ((P ∨ Q) ∧ (P ∨ R))
lemma
¬ (P = Q) ≠ ((Q ∨ ¬ P) ∧ (¬ Q ∨ P))
lemma
¬ (P --> Q) ≠ (¬ P ∨ Q)
lemma
¬ ¬ ((P --> Q) ∨ (Q --> P))
lemma
(P ∧ Q) = (P = (Q = (P ∨ Q)))
lemma
(∃x. P x --> Q) --> (∀x. P x) --> Q
lemma
((∀x. P x) --> Q) --> ¬ (∀x. P x ∧ ¬ Q)
lemma
((∀x. ¬ P x) --> Q) --> ¬ (∀x. ¬ (P x ∨ Q))
lemma
(∀x. P x) ∨ Q --> (∀x. P x ∨ Q)
lemma
(∃x. P --> Q x) --> P --> (∃x. Q x)
lemma
¬ ¬ (∃x. ∀y z. (P y --> Q z) --> P x --> Q x)
lemma
(∀x y. ∃z. ∀w. P x ∧ Q y --> R z ∧ S w) --> (∃x y. P x ∧ Q y) --> (∃z. R z)
lemma
(∃x. P --> Q x) ∧ (∃x. Q x --> P) --> ¬ ¬ (∃x. P = Q x)
lemma
(∀x. P = Q x) --> P = (∀x. Q x)
lemma
¬ (∀x. P ∨ Q x) ≠ (P ∨ (∀x. Q x))
lemma
(∃x. P x) ∧
(∀x. L x --> ¬ (M x ∧ R x)) ∧
(∀x. P x --> M x ∧ L x) ∧ ((∀x. P x --> Q x) ∨ (∃x. P x ∧ R x)) -->
(∃x. Q x ∧ P x)
lemma
(∃x. P x ∧ ¬ Q x) ∧
(∀x. P x --> R x) ∧
(∀x. M x ∧ L x --> P x) ∧ ((∃x. R x ∧ ¬ Q x) --> (∀x. L x --> ¬ R x)) -->
(∀x. M x --> ¬ L x)
lemma
(∀x. P x --> (∀x. Q x)) ∧
(¬ ¬ (∀x. Q x ∨ R x) --> (∃x. Q x ∧ S x)) ∧
(¬ ¬ (∃x. S x) --> (∀x. L x --> M x)) -->
(∀x. P x ∧ L x --> M x)
lemma
(∃x. P x) ∧ (∃y. Q y) -->
((∀x. P x --> R x) ∧ (∀y. Q y --> S y)) = (∀x y. P x ∧ Q y --> R x ∧ S y)
lemma
(∀x. P x ∨ Q x --> ¬ R x) ∧ (∀x. (Q x --> ¬ S x) --> P x ∧ R x) -->
(∀x. ¬ ¬ S x)
lemma
¬ (∃x. P x ∧ (Q x ∨ R x)) ∧ (∃x. L x ∧ P x) ∧ (∀x. ¬ R x --> M x) -->
(∃x. L x ∧ M x)
lemma
(∀x. P x ∧ (Q x ∨ R x) --> S x) ∧
(∀x. S x ∧ R x --> L x) ∧ (∀x. M x --> R x) -->
(∀x. P x ∧ M x --> L x)
lemma
(∀x. ∃y. J x y) ∧
(∀x. ∃y. G x y) ∧ (∀x y. J x y ∨ G x y --> (∀z. J y z ∨ G y z --> H x z)) -->
(∀x. ∃y. H x y)
lemma
¬ (∃x. ∀y. F y x = (¬ F y y))
lemma
(∃y. ∀x. F x y = F x x) --> ¬ (∀x. ∃y. ∀z. F z y = (¬ F z x))
lemma
(∀x. f x --> (∃y. g y ∧ h x y ∧ (∃y. g y ∧ ¬ h x y))) ∧
(∃x. j x ∧ (∀y. g y --> h x y)) -->
(∃x. j x ∧ ¬ f x)
lemma
(a = b ∨ c = d) ∧ (a = c ∨ b = d) --> a = d ∨ b = c
lemma
(∃z w. ∀x y. P x y = (x = z ∧ y = w)) -->
(∃z. ∀x. ∃w. (∀y. P x y = (y = w)) = (x = z))
lemma
(∃z w. ∀x y. P x y = (x = z ∧ y = w)) -->
(∃w. ∀y. ∃z. (∀x. P x y = (x = z)) = (y = w))
lemma
(∀x. (∃y. P y ∧ x = f y) --> P x) = (∀x. P x --> P (f x))
lemma
P (f a b) (f b c) ∧ P (f b c) (f a c) ∧ (∀x y z. P x y ∧ P y z --> P x z) -->
P (f a b) (f a c)
lemma
∀x. P x (f x) = (∃y. (∀z. P z y --> P z (f x)) ∧ P x y)