# Theory Intuitionistic

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theory Intuitionistic
imports Main
begin

    (*  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)