(* 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 \<open>Higher-Order Logic: Intuitionistic predicate calculus problems\<close>
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 "(\x. \y. \z. p(x) \ q(y) \ r(z)) =
(\<forall>z. \<exists>y. \<forall>x. p(x) \<and> q(y) \<and> r(z))"
by (iprover del: allE elim 2: allE')
(*Problem 3.1*)
lemma "\ (\x. \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 "(\x. P(x)\Q) \ (\x. P(x)) \ Q"
by iprover
lemma "((\x. P(x))\Q) \ \ (\x. P(x) \ \Q)"
by iprover
lemma "((\x. \P(x))\Q) \ \ (\x. \ (P(x)\Q))"
by iprover
lemma "(\x. P(x)) \ Q \ (\x. P(x) \ Q)"
by iprover
lemma "(\x. P \ Q(x)) \ (P \ (\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 "\\(\x. \y z. (P(y)\Q(z)) \ (P(x)\Q(x)))"
by iprover
(* Problem 20 *)
lemma "(\x y. \z. \w. (P(x)\Q(y)\R(z)\S(w)))
\<longrightarrow> (\<exists>x y. P(x) \<and> Q(y)) \<longrightarrow> (\<exists>z. R(z))"
by iprover
(* Problem 21 *)
lemma "(\x. P\Q(x)) \ (\x. Q(x)\P) \ \\(\x. P=Q(x))"
by iprover
(* Problem 22 *)
lemma "(\x. P = Q(x)) \ (P = (\x. Q(x)))"
by iprover
(* Problem ~~23 *)
lemma "\\ ((\x. P \ Q(x)) = (P \ (\x. Q(x))))"
by iprover
(* Problem 25 *)
lemma "(\x. P(x)) \
(\<forall>x. L(x) \<longrightarrow> \<not> (M(x) \<and> R(x))) \<and>
(\<forall>x. P(x) \<longrightarrow> (M(x) \<and> L(x))) \<and>
((\<forall>x. P(x)\<longrightarrow>Q(x)) \<or> (\<exists>x. P(x)\<and>R(x)))
\<longrightarrow> (\<exists>x. Q(x)\<and>P(x))"
by iprover
(* Problem 27 *)
lemma "(\x. P(x) \ \Q(x)) \
(\<forall>x. P(x) \<longrightarrow> R(x)) \<and>
(\<forall>x. M(x) \<and> L(x) \<longrightarrow> P(x)) \<and>
((\<exists>x. R(x) \<and> \<not> Q(x)) \<longrightarrow> (\<forall>x. L(x) \<longrightarrow> \<not> R(x)))
\<longrightarrow> (\<forall>x. M(x) \<longrightarrow> \<not>L(x))"
by iprover
(* Problem ~~28. AMENDED *)
lemma "(\x. P(x) \ (\x. Q(x))) \
(\<not>\<not>(\<forall>x. Q(x)\<or>R(x)) \<longrightarrow> (\<exists>x. Q(x)&S(x))) \<and>
(\<not>\<not>(\<exists>x. S(x)) \<longrightarrow> (\<forall>x. L(x) \<longrightarrow> M(x)))
\<longrightarrow> (\<forall>x. P(x) \<and> L(x) \<longrightarrow> M(x))"
by iprover
(* Problem 29. Essentially the same as Principia Mathematica *11.71 *)
lemma "(((\x. P(x)) \ (\y. Q(y))) \
(((\<forall>x. (P(x) \<longrightarrow> R(x))) \<and> (\<forall>y. (Q(y) \<longrightarrow> S(y)))) =
(\<forall>x y. ((P(x) \<and> Q(y)) \<longrightarrow> (R(x) \<and> S(y))))))"
by iprover
(* Problem ~~30 *)
lemma "(\x. (P(x) \ Q(x)) \ \ R(x)) \
(\<forall>x. (Q(x) \<longrightarrow> \<not> S(x)) \<longrightarrow> P(x) \<and> R(x))
\<longrightarrow> (\<forall>x. \<not>\<not>S(x))"
by iprover
(* Problem 31 *)
lemma "\(\x. P(x) \ (Q(x) \ R(x))) \
(\<exists>x. L(x) \<and> P(x)) \<and>
(\<forall>x. \<not> R(x) \<longrightarrow> M(x))
\<longrightarrow> (\<exists>x. L(x) \<and> M(x))"
by iprover
(* Problem 32 *)
lemma "(\x. P(x) \ (Q(x)|R(x))\S(x)) \
(\<forall>x. S(x) \<and> R(x) \<longrightarrow> L(x)) \<and>
(\<forall>x. M(x) \<longrightarrow> R(x))
\<longrightarrow> (\<forall>x. P(x) \<and> M(x) \<longrightarrow> L(x))"
by iprover
(* Problem ~~33 *)
lemma "(\x. \\(P(a) \ (P(x)\P(b))\P(c))) =
(\<forall>x. \<not>\<not>((\<not>P(a) \<or> P(x) \<or> P(c)) \<and> (\<not>P(a) \<or> \<not>P(b) \<or> P(c))))"
oops
(* Problem 36 *)
lemma
"(\x. \y. J x y) \
(\<forall>x. \<exists>y. G x y) \<and>
(\<forall>x y. J x y \<or> G x y \<longrightarrow> (\<forall>z. J y z \<or> G y z \<longrightarrow> H x z))
\<longrightarrow> (\<forall>x. \<exists>y. H x y)"
by iprover
(* Problem 39 *)
lemma "\ (\x. \y. F y x = (\F y y))"
by iprover
(* Problem 40. AMENDED *)
lemma "(\y. \x. F x y = F x x) \
\<not>(\<forall>x. \<exists>y. \<forall>z. F z y = (\<not> F z x))"
by iprover
(* Problem 44 *)
lemma "(\x. f(x) \
(\<exists>y. g(y) \<and> h x y \<and> (\<exists>y. g(y) \<and> ~ h x y))) \<and>
(\<exists>x. j(x) \<and> (\<forall>y. g(y) \<longrightarrow> h x y))
\<longrightarrow> (\<exists>x. j(x) \<and> \<not>f(x))"
by iprover
(* Problem 48 *)
lemma "(a=b \ c=d) \ (a=c \ b=d) \ a=d \ b=c"
by iprover
(* Problem 51 *)
lemma "((\z w. (\x y. (P x y = ((x = z) \ (y = w))))) \
(\<exists>z. (\<forall>x. (\<exists>w. ((\<forall>y. (P x y = (y = w))) = (x = z))))))"
by iprover
(* Problem 52 *)
(*Almost the same as 51. *)
lemma "((\z w. (\x y. (P x y = ((x = z) \ (y = w))))) \
(\<exists>w. (\<forall>y. (\<exists>z. ((\<forall>x. (P x y = (x = z))) = (y = w))))))"
by iprover
(* Problem 56 *)
lemma "(\x. (\y. P(y) \ x=f(y)) \ P(x)) = (\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) \
(\<forall>x y z. P x y \<and> P y z \<longrightarrow> P x z) \<longrightarrow> P (f a b) (f a c)"
by iprover
(* Problem 60 *)
lemma "\x. P x (f x) = (\y. (\z. P z y \ P z (f x)) \ P x y)"
by iprover
end
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