TPTP Problem File: SYO528^1.p
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% File : SYO528^1 : TPTP v9.0.0. Released v5.2.0.
% Domain : Syntactic
% Problem : There can be 4 distinct choice operators on type $o
% Version : Especial.
% English :
% Refs : [Bro11] Brown E. (2011), Email to Geoff Sutcliffe
% Source : [Bro11]
% Names : CHOICE5 [Bro11]
% Status : Satisfiable
% Rating : 0.33 v8.2.0, 0.00 v8.1.0, 0.33 v6.1.0, 0.00 v5.4.0, 0.67 v5.2.0
% Syntax : Number of formulae : 14 ( 10 unt; 4 typ; 0 def)
% Number of atoms : 22 ( 6 equ; 0 cnn)
% Maximal formula atoms : 1 ( 2 avg)
% Number of connectives : 22 ( 6 ~; 0 |; 0 &; 12 @)
% ( 0 <=>; 4 =>; 0 <=; 0 <~>)
% Maximal formula depth : 5 ( 3 avg)
% Number of types : 1 ( 0 usr)
% Number of type conns : 12 ( 12 >; 0 *; 0 +; 0 <<)
% Number of symbols : 5 ( 4 usr; 0 con; 1-2 aty)
% Number of variables : 8 ( 0 ^; 4 !; 4 ?; 8 :)
% SPC : TH0_SAT_EQU_NAR
% Comments :
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thf(eps1,type,
eps1: ( $o > $o ) > $o ).
thf(choiceax1,axiom,
! [P: $o > $o] :
( ? [X: $o] : ( P @ X )
=> ( P @ ( eps1 @ P ) ) ) ).
thf(eps2,type,
eps2: ( $o > $o ) > $o ).
thf(choiceax2,axiom,
! [P: $o > $o] :
( ? [X: $o] : ( P @ X )
=> ( P @ ( eps2 @ P ) ) ) ).
thf(eps3,type,
eps3: ( $o > $o ) > $o ).
thf(choiceax3,axiom,
! [P: $o > $o] :
( ? [X: $o] : ( P @ X )
=> ( P @ ( eps3 @ P ) ) ) ).
thf(eps4,type,
eps4: ( $o > $o ) > $o ).
thf(choiceax4,axiom,
! [P: $o > $o] :
( ? [X: $o] : ( P @ X )
=> ( P @ ( eps4 @ P ) ) ) ).
thf(choiceax12,axiom,
eps1 != eps2 ).
thf(choiceax13,axiom,
eps1 != eps3 ).
thf(choiceax14,axiom,
eps1 != eps4 ).
thf(choiceax23,axiom,
eps2 != eps3 ).
thf(choiceax24,axiom,
eps2 != eps4 ).
thf(choiceax34,axiom,
eps3 != eps4 ).
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