TPTP Problem File: SWC072+1.p
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%--------------------------------------------------------------------------
% File : SWC072+1 : TPTP v9.0.0. Released v2.4.0.
% Domain : Software Creation
% Problem : cond_id_segment_total2_x_rot_l_total1
% Version : [Wei00] axioms.
% English : Find components in a software library that match a given target
% specification given in first-order logic. The components are
% specified in first-order logic as well. The problem represents
% a test of one library module specification against a target
% specification.
% Refs : [Wei00] Weidenbach (2000), Software Reuse of List Functions Ve
% : [FSS98] Fischer et al. (1998), Deduction-Based Software Compon
% Source : [Wei00]
% Names : cond_id_segment_total2_x_rot_l_total1 [Wei00]
% Status : Theorem
% Rating : 0.76 v9.0.0, 0.78 v8.2.0, 0.72 v8.1.0, 0.75 v7.5.0, 0.81 v7.4.0, 0.73 v7.3.0, 0.83 v7.1.0, 0.78 v7.0.0, 0.77 v6.3.0, 0.79 v6.2.0, 0.92 v6.1.0, 0.93 v6.0.0, 0.96 v5.3.0, 1.00 v5.2.0, 0.95 v5.0.0, 0.96 v4.1.0, 1.00 v2.4.0
% Syntax : Number of formulae : 96 ( 9 unt; 0 def)
% Number of atoms : 420 ( 82 equ)
% Maximal formula atoms : 26 ( 4 avg)
% Number of connectives : 353 ( 29 ~; 14 |; 54 &)
% ( 26 <=>; 230 =>; 0 <=; 0 <~>)
% Maximal formula depth : 29 ( 7 avg)
% Maximal term depth : 4 ( 1 avg)
% Number of predicates : 20 ( 19 usr; 0 prp; 1-2 aty)
% Number of functors : 5 ( 5 usr; 1 con; 0-2 aty)
% Number of variables : 212 ( 194 !; 18 ?)
% SPC : FOF_THM_RFO_SEQ
% Comments :
%--------------------------------------------------------------------------
%----Include list specification axioms
include('Axioms/SWC001+0.ax').
%--------------------------------------------------------------------------
fof(co1,conjecture,
! [U] :
( ssList(U)
=> ! [V] :
( ssList(V)
=> ! [W] :
( ssList(W)
=> ! [X] :
( ~ ssList(X)
| V != X
| U != W
| ? [Y] :
( ssList(Y)
& neq(Y,nil)
& segmentP(V,Y)
& segmentP(U,Y) )
| ( nil != W
& nil = X )
| ( nil = V
& nil = U )
| ( ? [Z] :
( ssList(Z)
& W != Z
& ? [X1] :
( ssList(X1)
& ? [X2] :
( ssList(X2)
& tl(X) = X1
& app(X1,X2) = Z
& ? [X3] :
( ssItem(X3)
& cons(X3,nil) = X2
& hd(X) = X3
& neq(nil,X) )
& neq(nil,X) ) ) )
& neq(X,nil) ) ) ) ) ) ).
%--------------------------------------------------------------------------