TPTP Problem File: SET745+4.p
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- Solve Problem
%--------------------------------------------------------------------------
% File : SET745+4 : TPTP v9.0.0. Bugfixed v2.2.1.
% Domain : Set Theory (Mappings)
% Problem : Problem on composition of mappings 10
% Version : [Pas99] axioms.
% English : Consider three mappings F1 from A1 to B,F2 from A2 to B,
% F which is equal to F1 on A1 and to F2 on A2, then F is a
% mapping from union(A1,A2) to B if and only if F1 and F2 are
% equal on the intersection of A1 and A2.
% Refs : [Pas99] Pastre (1999), Email to G. Sutcliffe
% Source : [Pas99]
% Names :
% Status : Theorem
% Rating : 0.76 v9.0.0, 0.75 v8.2.0, 0.78 v7.5.0, 0.75 v7.4.0, 0.67 v7.3.0, 0.66 v7.1.0, 0.61 v7.0.0, 0.73 v6.4.0, 0.77 v6.3.0, 0.71 v6.2.0, 0.84 v6.1.0, 0.83 v6.0.0, 0.87 v5.5.0, 0.89 v5.4.0, 0.93 v5.2.0, 0.95 v5.0.0, 0.96 v3.7.0, 0.95 v3.3.0, 0.93 v3.2.0, 0.91 v3.1.0, 0.89 v2.7.0, 0.83 v2.6.0, 0.86 v2.5.0, 0.88 v2.4.0, 0.75 v2.3.0, 0.67 v2.2.1
% Syntax : Number of formulae : 29 ( 1 unt; 0 def)
% Number of atoms : 145 ( 7 equ)
% Maximal formula atoms : 17 ( 5 avg)
% Number of connectives : 118 ( 2 ~; 3 |; 60 &)
% ( 32 <=>; 21 =>; 0 <=; 0 <~>)
% Maximal formula depth : 19 ( 9 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 16 ( 15 usr; 0 prp; 2-6 aty)
% Number of functors : 15 ( 15 usr; 1 con; 0-5 aty)
% Number of variables : 144 ( 135 !; 9 ?)
% SPC : FOF_THM_RFO_SEQ
% Comments :
% Bugfixes : v2.2.1 - Bugfixes in SET006+1.ax.
%--------------------------------------------------------------------------
%----Include set theory definitions
include('Axioms/SET006+0.ax').
%----Include mappings axioms
include('Axioms/SET006+1.ax').
%--------------------------------------------------------------------------
fof(thII36,conjecture,
! [F1,F2,F,A1,A2,B] :
( ( maps(F1,A1,B)
& maps(F2,A2,B)
& ! [X,Y] :
( ( member(X,union(A1,A2))
& member(Y,B) )
=> ( apply(F,X,Y)
<=> ( ( member(X,A1)
& apply(F1,X,Y) )
| ( member(X,A2)
& apply(F2,X,Y) ) ) ) ) )
=> ( maps(F,union(A1,A2),B)
<=> ! [X,Y1,Y2] :
( ( member(X,A1)
& member(X,A2)
& member(Y1,B)
& member(Y2,B)
& apply(F1,X,Y1)
& apply(F2,X,Y2) )
=> Y1 = Y2 ) ) ) ).
%--------------------------------------------------------------------------