(* Title: HOL/ex/Intuitionistic.thy
ID: $Id$
Author: Lawrence C Paulson, Cambridge University Computer Laboratory
Copyright 1991 University of Cambridge
Higher-Order Logic: Intuitionistic predicate calculus problems
Taken from FOL/ex/int.ML
*)
theory Intuitionistic = Main:
(*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 rules
lemma "~~ ((~P --> Q) --> (~P --> ~Q) --> P)"
by rules
(* ~~ does NOT distribute over | *)
lemma "(~~(P-->Q)) = (~~P --> ~~Q)"
by rules
lemma "(~~~P) = (~P)"
by rules
lemma "~~((P --> Q | R) --> (P-->Q) | (P-->R))"
by rules
lemma "(P=Q) = (Q=P)"
by rules
lemma "((P --> (Q | (Q-->R))) --> R) --> R"
by rules
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 rules
(* Lemmas for the propositional double-negation translation *)
lemma "P --> ~~P"
by rules
lemma "~~(~~P --> P)"
by rules
lemma "~~P & ~~(P --> Q) --> ~~Q"
by rules
(* 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 rules
(*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 rules
(*** 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 (rules del: allE elim 2: allE')
(*Problem 3.1*)
lemma "~ (EX x. ALL y. p y x = (~ p x x))"
by rules
(* Intuitionistic FOL: propositional problems based on Pelletier. *)
(* Problem ~~1 *)
lemma "~~((P-->Q) = (~Q --> ~P))"
by rules
(* Problem ~~2 *)
lemma "~~(~~P = P)"
by rules
(* Problem 3 *)
lemma "~(P-->Q) --> (Q-->P)"
by rules
(* Problem ~~4 *)
lemma "~~((~P-->Q) = (~Q --> P))"
by rules
(* Problem ~~5 *)
lemma "~~((P|Q-->P|R) --> P|(Q-->R))"
by rules
(* Problem ~~6 *)
lemma "~~(P | ~P)"
by rules
(* Problem ~~7 *)
lemma "~~(P | ~~~P)"
by rules
(* Problem ~~8. Peirce's law *)
lemma "~~(((P-->Q) --> P) --> P)"
by rules
(* Problem 9 *)
lemma "((P|Q) & (~P|Q) & (P| ~Q)) --> ~ (~P | ~Q)"
by rules
(* Problem 10 *)
lemma "(Q-->R) --> (R-->P&Q) --> (P-->(Q|R)) --> (P=Q)"
by rules
(* 11. Proved in each direction (incorrectly, says Pelletier!!) *)
lemma "P=P"
by rules
(* Problem ~~12. Dijkstra's law *)
lemma "~~(((P = Q) = R) = (P = (Q = R)))"
by rules
lemma "((P = Q) = R) --> ~~(P = (Q = R))"
by rules
(* Problem 13. Distributive law *)
lemma "(P | (Q & R)) = ((P | Q) & (P | R))"
by rules
(* Problem ~~14 *)
lemma "~~((P = Q) = ((Q | ~P) & (~Q|P)))"
by rules
(* Problem ~~15 *)
lemma "~~((P --> Q) = (~P | Q))"
by rules
(* Problem ~~16 *)
lemma "~~((P-->Q) | (Q-->P))"
by rules
(* 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 rules
(****Examples with quantifiers****)
(* The converse is classical in the following implications... *)
lemma "(EX x. P(x)-->Q) --> (ALL x. P(x)) --> Q"
by rules
lemma "((ALL x. P(x))-->Q) --> ~ (ALL x. P(x) & ~Q)"
by rules
lemma "((ALL x. ~P(x))-->Q) --> ~ (ALL x. ~ (P(x)|Q))"
by rules
lemma "(ALL x. P(x)) | Q --> (ALL x. P(x) | Q)"
by rules
lemma "(EX x. P --> Q(x)) --> (P --> (EX x. Q(x)))"
by rules
(* 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 rules
(* 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 rules
(* Problem 21 *)
lemma "(EX x. P-->Q(x)) & (EX x. Q(x)-->P) --> ~~(EX x. P=Q(x))"
by rules
(* Problem 22 *)
lemma "(ALL x. P = Q(x)) --> (P = (ALL x. Q(x)))"
by rules
(* Problem ~~23 *)
lemma "~~ ((ALL x. P | Q(x)) = (P | (ALL x. Q(x))))"
by rules
(* 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 rules
(* 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 rules
(* 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 rules
(* 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 rules
(* 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 rules
(* 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 rules
(* 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 rules
(* 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 rules
(* Problem 39 *)
lemma "~ (EX x. ALL y. F y x = (~F y y))"
by rules
(* 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 rules
(* 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 rules
(* Problem 48 *)
lemma "(a=b | c=d) & (a=c | b=d) --> a=d | b=c"
by rules
(* 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 rules
(* 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 rules
(* Problem 56 *)
lemma "(ALL x. (EX y. P(y) & x=f(y)) --> P(x)) = (ALL x. P(x) --> P(f(x)))"
by rules
(* 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 rules
(* Problem 60 *)
lemma "ALL x. P x (f x) = (EX y. (ALL z. P z y --> P z (f x)) & P x y)"
by rules
end