TPTP Problem File: SEV191^5.p
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- Solve Problem
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% File : SEV191^5 : TPTP v9.0.0. Released v4.0.0.
% Domain : Set Theory (Sets of sets)
% Problem : TPS problem S-JOINFN-MONOTONE
% Version : Especial.
% English : The function used to define JOIN as a LFP is monotone.
% Refs : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source : [Bro09]
% Names : tps_0503 [Bro09]
% : S-JOINFN-MONOTONE [TPS]
% Status : Theorem
% Rating : 0.12 v9.0.0, 0.20 v8.2.0, 0.15 v8.1.0, 0.18 v7.5.0, 0.14 v7.4.0, 0.22 v7.2.0, 0.12 v7.1.0, 0.25 v7.0.0, 0.29 v6.4.0, 0.33 v6.3.0, 0.40 v6.2.0, 0.29 v5.5.0, 0.17 v5.4.0, 0.20 v5.1.0, 0.40 v5.0.0, 0.20 v4.1.0, 0.00 v4.0.0
% Syntax : Number of formulae : 4 ( 0 unt; 3 typ; 0 def)
% Number of atoms : 18 ( 14 equ; 0 cnn)
% Maximal formula atoms : 18 ( 18 avg)
% Number of connectives : 53 ( 0 ~; 4 |; 15 &; 30 @)
% ( 0 <=>; 4 =>; 0 <=; 0 <~>)
% Maximal formula depth : 24 ( 24 avg)
% Number of types : 2 ( 1 usr)
% Number of type conns : 11 ( 11 >; 0 *; 0 +; 0 <<)
% Number of symbols : 4 ( 2 usr; 2 con; 0-2 aty)
% Number of variables : 21 ( 0 ^; 9 !; 12 ?; 21 :)
% 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/
% : Polymorphic definitions expanded.
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thf(a_type,type,
a: $tType ).
thf(cP,type,
cP: a > a > a ).
thf(c0,type,
c0: a ).
thf(cS_JOINFN_MONOTONE_pme,conjecture,
( ! [R: a > a > a > $o] :
( $true
=> $true )
& ! [R: a > a > a > $o,S: a > a > a > $o] :
( ( $true
& $true
& ! [Xa: a,Xb: a,Xc: a] :
( ( R @ Xa @ Xb @ Xc )
=> ( S @ Xa @ Xb @ Xc ) ) )
=> ! [Xa: a,Xb: a,Xc: a] :
( ( ( ( Xa = c0 )
& ( Xb = Xc ) )
| ( ( Xb = c0 )
& ( Xa = Xc ) )
| ? [Xx1: a,Xx2: a,Xy1: a,Xy2: a,Xz1: a,Xz2: a] :
( ( Xa
= ( cP @ Xx1 @ Xx2 ) )
& ( Xb
= ( cP @ Xy1 @ Xy2 ) )
& ( Xc
= ( cP @ Xz1 @ Xz2 ) )
& ( R @ Xx1 @ Xy1 @ Xz1 )
& ( R @ Xx2 @ Xy2 @ Xz2 ) ) )
=> ( ( ( Xa = c0 )
& ( Xb = Xc ) )
| ( ( Xb = c0 )
& ( Xa = Xc ) )
| ? [Xx1: a,Xx2: a,Xy1: a,Xy2: a,Xz1: a,Xz2: a] :
( ( Xa
= ( cP @ Xx1 @ Xx2 ) )
& ( Xb
= ( cP @ Xy1 @ Xy2 ) )
& ( Xc
= ( cP @ Xz1 @ Xz2 ) )
& ( S @ Xx1 @ Xy1 @ Xz1 )
& ( S @ Xx2 @ Xy2 @ Xz2 ) ) ) ) ) ) ).
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