TSTP Solution File: ROB015-1 by DarwinFM---1.4.5

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : DarwinFM---1.4.5
% Problem  : ROB015-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : darwin -fd true -ppp true -pl 0 -to %d -pmtptp true %s

% Computer : n023.cluster.edu
% Model    : x86_64 x86_64
% CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory   : 8042.1875MB
% OS       : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit  : 0s
% DateTime : Mon Jul 18 20:50:54 EDT 2022

% Result   : Satisfiable 0.19s 0.42s
% Output   : FiniteModel 0.19s
% Verified : 
% SZS Type : FiniteModel
%            Domain size           :    2

% Comments : 
%------------------------------------------------------------------------------
fof(interpretation_domain,fi_domain,
    ! [X] :
      ( X = e1
      | X = e2 ) ).

fof(interpretation_domain_distinct,fi_domain,
    e1 != e2 ).

fof(interpretation_terms,fi_functors,
    ( ! [X0,X1,X2] :
        ( add(X0,X1) = X2
      <=> ( ( X2 = e1
            & X0 != e2
            & X1 != e2 )
          | ( X1 = e1
            & X2 = e1
            & X0 != e2 )
          | ( X1 = e2
            & X2 = e2 )
          | ( X0 = e2
            & X2 = e2 ) ) )
    & d = e1
    & e = e1
    & k = e1
    & ! [X0,X1,X2] :
        ( multiply(X0,X1) = X2
      <=> ( ( X2 = e1
            & ~ ( X0 = e2
                & X1 = e2 )
            & ~ ( X0 = e1
                & X1 = e2 )
            & ~ ( X0 = e1
                & X1 = e1 ) )
          | ( X0 = e1
            & X1 = e1
            & X2 = e2 )
          | ( X0 = e1
            & X1 = e2
            & X2 = e2 )
          | ( X0 = e2
            & X1 = e2
            & X2 = e2 ) ) )
    & ! [X0,X1] :
        ( negate(X0) = X1
      <=> ( ( X1 = e2
            & X0 != e2 )
          | ( X0 = e2
            & X1 = e1 ) ) )
    & one = e2
    & ! [X0,X1] :
        ( successor(X0) = X1
      <=> ( ( X1 = e2
            & X0 != e1 )
          | ( X0 = e1
            & X1 = e1 ) ) ) ) ).

fof(interpretation_atoms,fi_predicates,
    ! [X0] :
      ( positive_integer(X0)
    <=> ( X0 = e2
        | X0 = e1 ) ) ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12  % Problem  : ROB015-1 : TPTP v8.1.0. Released v1.0.0.
% 0.04/0.13  % Command  : darwin -fd true -ppp true -pl 0 -to %d -pmtptp true %s
% 0.13/0.34  % Computer : n023.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % DateTime : Thu Jun  9 13:40:52 EDT 2022
% 0.13/0.34  % CPUTime  : 
% 0.13/0.34  Defaulting to tptp format.
% 0.19/0.42  SZS status Satisfiable for /export/starexec/sandbox2/benchmark/theBenchmark.p
% 0.19/0.42  
% 0.19/0.42  MODEL (TPTP):
% 0.19/0.42  SZS output start FiniteModel for /export/starexec/sandbox2/benchmark/theBenchmark.p
% See solution above
%------------------------------------------------------------------------------