src/HOL/ex/Intuitionistic.thy

no longer declare .psimps rules as [simp].

This regularly caused confusion (e.g., they show up in simp traces
when the regular simp rules are disabled). In the few places where the
rules are used, explicitly mentioning them actually clarifies the
proof text.

(*  Title:      HOL/ex/Intuitionistic.thy
    ID:         $Id$
    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