TPTP Problem File: SYO558^1.p
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% File : SYO558^1 : TPTP v7.1.0. Released v5.2.0.
% Domain : Syntactic
% Problem : Forall on $i can be expressed in terms of choice on $
% Version : Especial.
% English :
% Refs : [Bro11] Brown (2011), Email to Geoff Sutcliffe
% Source : [Bro11]
% Names : CHOICE30 [Bro11]
% Status : Theorem
% Rating : 0.25 v7.1.0, 0.38 v7.0.0, 0.29 v6.4.0, 0.33 v6.3.0, 0.40 v6.2.0, 0.57 v6.1.0, 0.43 v5.5.0, 0.50 v5.4.0, 0.80 v5.2.0
% Syntax : Number of formulae : 3 ( 0 unit; 1 type; 0 defn)
% Number of atoms : 11 ( 1 equality; 8 variable)
% Maximal formula depth : 8 ( 6 average)
% Number of connectives : 9 ( 1 ~; 0 |; 0 &; 6 @)
% ( 0 <=>; 1 =>; 0 <=; 0 <~>)
% ( 0 ~|; 0 ~&)
% Number of type conns : 5 ( 5 >; 0 *; 0 +; 0 <<)
% Number of symbols : 3 ( 1 :; 0 =)
% Number of variables : 5 ( 0 sgn; 1 !; 1 ?; 3 ^)
% ( 5 :; 0 !>; 0 ?*)
% ( 0 @-; 0 @+)
% SPC : TH0_THM_EQU_NAR
% Comments : (eps (complement P)) can only satisfy P if everything satisfies P.
% Hence (^P.P(eps (^x.~Px))) = (^P.!! P).
% : Note that this problem uses !! for forall instead of the !
% quantifier.
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thf(eps,type,(
eps: ( $i > $o ) > $i )).
thf(choiceax,axiom,(
! [P: $i > $o] :
( ? [X: $i] :
( P @ X )
=> ( P @ ( eps @ P ) ) ) )).
thf(conj,conjecture,
( ( ^ [P: $i > $o] :
( P
@ ( eps
@ ^ [X: $i] :
~ ( P @ X ) ) ) )
= ( ^ [P: $i > $o] :
!! ( P ) ) )).
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