TPTP Problem File: PUZ127^5.p
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% File : PUZ127^5 : TPTP v9.0.0. Released v4.0.0.
% Domain : Puzzles
% Problem : TPS problem from CHECKERBOARD-THMS
% Version : Especial.
% English :
% Refs : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source : [Bro09]
% Names : tps_1182 [Bro09]
% Status : Theorem
% Rating : 0.75 v9.0.0, 0.70 v8.2.0, 0.62 v8.1.0, 0.64 v7.5.0, 0.71 v7.4.0, 0.67 v7.2.0, 0.62 v7.1.0, 0.75 v7.0.0, 0.71 v6.4.0, 0.83 v6.3.0, 0.80 v6.2.0, 0.71 v5.5.0, 0.67 v5.4.0, 1.00 v4.0.0
% Syntax : Number of formulae : 9 ( 0 unt; 8 typ; 0 def)
% Number of atoms : 25 ( 25 equ; 0 cnn)
% Maximal formula atoms : 25 ( 25 avg)
% Number of connectives : 77 ( 6 ~; 5 |; 16 &; 47 @)
% ( 3 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 19 ( 19 avg)
% Number of types : 1 ( 0 usr)
% Number of type conns : 3 ( 3 >; 0 *; 0 +; 0 <<)
% Number of symbols : 9 ( 8 usr; 6 con; 0-2 aty)
% Number of variables : 11 ( 0 ^; 11 !; 0 ?; 11 :)
% SPC : TH0_THM_EQU_NAR
% Comments : This problem is from the TPS library. Copyright (c) 2009 The TPS
% project in the Department of Mathematical Sciences at Carnegie
% Mellon University. Distributed under the Creative Commons copyleft
% license: http://creativecommons.org/licenses/by-sa/3.0/
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thf(c5,type,
c5: $i ).
thf(g,type,
g: $i > $i > $i ).
thf(c4,type,
c4: $i ).
thf(c3,type,
c3: $i ).
thf(c2,type,
c2: $i ).
thf(c1,type,
c1: $i ).
thf(s,type,
s: $i > $i ).
thf(c8,type,
c8: $i ).
thf(cTOUGHNUT2,conjecture,
~ ( ( ( s @ ( s @ ( s @ ( s @ ( s @ ( s @ ( s @ ( s @ c8 ) ) ) ) ) ) ) )
= c8 )
& ! [Xx: $i] :
( ( s @ ( s @ ( s @ ( s @ Xx ) ) ) )
!= Xx )
& ! [Xx: $i,Xy: $i] :
( ( ( g @ Xx @ Xy )
= c5 )
<=> ( ( ( Xx = c8 )
& ( Xy = c8 ) )
| ( ( Xx = c1 )
& ( Xy = c1 ) ) ) )
& ! [Xx: $i,Xy: $i] :
( ( ( g @ Xx @ Xy )
= c1 )
<=> ( ( g @ ( s @ Xx ) @ Xy )
= c3 ) )
& ! [Xx: $i,Xy: $i] :
( ( ( g @ Xx @ Xy )
= c2 )
<=> ( ( g @ Xx @ ( s @ Xy ) )
= c4 ) )
& ! [Xx: $i,Xy: $i] :
( ( ( g @ c1 @ Xy )
!= c3 )
& ( ( g @ c8 @ Xy )
!= c1 )
& ( ( g @ Xx @ c1 )
!= c4 )
& ( ( g @ Xx @ c8 )
!= c2 ) )
& ( c1
= ( s @ c8 ) )
& ( c2
= ( s @ c1 ) )
& ( c3
= ( s @ c2 ) )
& ( c4
= ( s @ c3 ) )
& ( c5
= ( s @ c4 ) )
& ! [Xx: $i,Xy: $i] :
( ( ( g @ Xx @ Xy )
= c1 )
| ( ( g @ Xx @ Xy )
= c2 )
| ( ( g @ Xx @ Xy )
= c3 )
| ( ( g @ Xx @ Xy )
= c4 )
| ( ( g @ Xx @ Xy )
= c5 ) ) ) ).
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