TPTP Problem File: SWW096+1.p
View Solutions
- Solve Problem
%------------------------------------------------------------------------------
% File : SWW096+1 : TPTP v9.0.0. Released v5.2.0.
% Domain : Software Verification
% Problem : Equivalenace of the semantic and syntactic definition of and
% Version : [deN09] axioms.
% English :
% Refs : [deN09] de Nivelle (2009), Email to Geoff Sutcliffe
% : [deN10] de Nivelle (2010), Classical Logic with Partial Functi
% Source : [deN09]
% Names :
% Status : Theorem
% Rating : 0.79 v9.0.0, 0.83 v8.2.0, 0.81 v8.1.0, 0.75 v7.5.0, 0.81 v7.4.0, 0.77 v7.3.0, 0.83 v7.1.0, 0.78 v7.0.0, 0.70 v6.4.0, 0.73 v6.3.0, 0.79 v6.2.0, 0.88 v6.1.0, 0.93 v6.0.0, 1.00 v5.2.0
% Syntax : Number of formulae : 45 ( 15 unt; 0 def)
% Number of atoms : 97 ( 50 equ)
% Maximal formula atoms : 9 ( 2 avg)
% Number of connectives : 77 ( 25 ~; 6 |; 26 &)
% ( 5 <=>; 15 =>; 0 <=; 0 <~>)
% Maximal formula depth : 10 ( 4 avg)
% Maximal term depth : 5 ( 1 avg)
% Number of predicates : 5 ( 4 usr; 0 prp; 1-2 aty)
% Number of functors : 27 ( 27 usr; 5 con; 0-3 aty)
% Number of variables : 77 ( 63 !; 14 ?)
% SPC : FOF_THM_RFO_SEQ
% Comments : They even agree on non truth values.
%------------------------------------------------------------------------------
%----Include axioms for Syntactic definitions of the logical operators
include('Axioms/SWV012+0.ax').
%------------------------------------------------------------------------------
fof(and1_and2,conjecture,
! [P,Q] : and1(P,Q) = and2(P,Q) ).
%------------------------------------------------------------------------------