(* Title: HOL/ex/Intuitionistic.thy Author: Lawrence C Paulson, Cambridge University Computer Laboratory Copyright 1991 University of Cambridge Taken from FOL/ex/int.ML *) section ‹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