# Theory Intuitionistic

theory Intuitionistic
imports Main
```(*  Title:      HOL/ex/Intuitionistic.thy
Author:     Lawrence C Paulson, Cambridge University Computer Laboratory

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 "(∀x. ∃y. ∀z. p(x) ∧ q(y) ∧ r(z)) =
(∀z. ∃y. ∀x. p(x) ∧ q(y) ∧ 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)))
⟶ (∃x y. P(x) ∧ Q(y)) ⟶ (∃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)) ∧
(∀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))"
by iprover

(* Problem 27 *)
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))"
by iprover

(* Problem ~~28.  AMENDED *)
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))"
by iprover

(* Problem 29.  Essentially the same as Principia Mathematica *11.71 *)
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))))))"
by iprover

(* Problem ~~30 *)
lemma "(∀x. (P(x) ∨ Q(x)) ⟶ ¬ R(x)) ∧
(∀x. (Q(x) ⟶ ¬ S(x)) ⟶ P(x) ∧ R(x))
⟶ (∀x. ¬¬S(x))"
by iprover

(* Problem 31 *)
lemma "¬(∃x. P(x) ∧ (Q(x) ∨ R(x))) ∧
(∃x. L(x) ∧ P(x)) ∧
(∀x. ¬ R(x) ⟶ M(x))
⟶ (∃x. L(x) ∧ M(x))"
by iprover

(* Problem 32 *)
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))"
by iprover

(* Problem ~~33 *)
lemma "(∀x. ¬¬(P(a) ∧ (P(x)⟶P(b))⟶P(c)))  =
(∀x. ¬¬((¬P(a) ∨ P(x) ∨ P(c)) ∧ (¬P(a) ∨ ¬P(b) ∨ P(c))))"
oops

(* Problem 36 *)
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)"
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) ⟶
¬(∀x. ∃y. ∀z. F z y = (¬ F z x))"
by iprover

(* Problem 44 *)
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))"
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))))) ⟶
(∃z. (∀x. (∃w. ((∀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))))) ⟶
(∃w. (∀y. (∃z. ((∀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) ∧
(∀x y z. P x y ∧ P y z ⟶ P x z) ⟶ 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
```