TPTP Problem File: SYO512^1.p
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% File : SYO512^1 : TPTP v8.2.0. Released v4.1.0.
% Domain : Syntactic
% Problem : Choice operator used to obtain functions from total relations
% Version : Especial.
% English : A choice operator can be used to obtain functions from total
% relations.
% Refs : [Bro09] Brown E. (2009), Email to Geoff Sutcliffe
% : [Hoe09] Hoeschele (2009), Towards a Semi-Automatic Higher-Orde
% Source : [Bro09]
% Names : choiceiskolem [Bro09]
% Status : Theorem
% Rating : 0.00 v8.1.0, 0.08 v7.4.0, 0.00 v6.2.0, 0.17 v6.1.0, 0.00 v4.1.0
% Syntax : Number of formulae : 5 ( 1 unt; 2 typ; 0 def)
% Number of atoms : 3 ( 0 equ; 0 cnn)
% Maximal formula atoms : 2 ( 1 avg)
% Number of connectives : 10 ( 0 ~; 0 |; 0 &; 9 @)
% ( 0 <=>; 1 =>; 0 <=; 0 <~>)
% Maximal formula depth : 5 ( 5 avg)
% Number of types : 2 ( 0 usr)
% Number of type conns : 5 ( 5 >; 0 *; 0 +; 0 <<)
% Number of symbols : 2 ( 2 usr; 0 con; 1-2 aty)
% Number of variables : 5 ( 0 ^; 3 !; 2 ?; 5 :)
% SPC : TH0_THM_NEQ_NAR
% Comments : I have given a choice operator eps for type i and assumed it is a
% choice operator. This is to get around using the choice operator
% @+ in TH0 syntax (since it is not ). eps maybe
% semantically different from the choice operator @+ but a theorem
% prover may recognize eps is a choice operator and treat it
% accordingly.
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thf(eps,type,
eps: ( $i > $o ) > $i ).
thf(epschoice,axiom,
! [P: $i > $o] :
( ? [X: $i] : ( P @ X )
=> ( P @ ( eps @ P ) ) ) ).
thf(r,type,
r: $i > $i > $o ).
thf(rtotal,axiom,
! [X: $i] :
? [Y: $i] : ( r @ X @ Y ) ).
thf(claim,conjecture,
! [X: $i] : ( r @ X @ ( eps @ ( r @ X ) ) ) ).
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