TSTP Solution File: LCL655+1.001 by Crossbow---0.1

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Crossbow---0.1
% Problem  : LCL655+1.001 : TPTP v8.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : do_Crossbow---0.1 %s

% Computer : n029.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  : 600s
% DateTime : Sun Jul 17 08:05:00 EDT 2022

% Result   : CounterSatisfiable 5.22s 5.39s
% Output   : FiniteModel 5.22s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem    : LCL655+1.001 : TPTP v8.1.0. Released v4.0.0.
% 0.03/0.12  % Command    : do_Crossbow---0.1 %s
% 0.12/0.33  % Computer : n029.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit   : 300
% 0.12/0.33  % WCLimit    : 600
% 0.12/0.33  % DateTime   : Sun Jul  3 12:34:25 EDT 2022
% 0.12/0.33  % CPUTime    : 
% 0.12/0.34  /export/starexec/sandbox/solver/bin
% 0.12/0.34  crossbow.opt
% 0.12/0.34  do_Crossbow---0.1
% 0.12/0.34  eprover
% 0.12/0.34  runsolver
% 0.12/0.34  starexec_run_Crossbow---0.1
% 5.22/5.39  % SZS status CounterSatisfiable for theBenchmark.p
% 5.22/5.39  % SZS output start FiniteModel for theBenchmark.p
% 5.22/5.39  % domain size: 2
% 5.22/5.39  fof(interp, fi_domain, ![X] : (X = 0 | X = 1)).
% 5.22/5.39  fof(interp, fi_functors, esk10_0 = 0).
% 5.22/5.39  fof(interp, fi_functors, esk11_0 = 1).
% 5.22/5.39  fof(interp, fi_functors, esk12_0 = 0).
% 5.22/5.39  fof(interp, fi_functors, esk1_0 = 0).
% 5.22/5.39  fof(interp, fi_functors, esk2_3(0, 0, 0) = 0 & esk2_3(0, 0, 1) = 0 &
% 5.22/5.39    esk2_3(0, 1, 0) = 0 &
% 5.22/5.39    esk2_3(0, 1, 1) = 0 &
% 5.22/5.39    esk2_3(1, 0, 0) = 0 &
% 5.22/5.39    esk2_3(1, 0, 1) = 0 &
% 5.22/5.39    esk2_3(1, 1, 0) = 0 &
% 5.22/5.39    esk2_3(1, 1, 1) = 0).
% 5.22/5.39  fof(interp, fi_functors, esk3_1(0) = 0 & esk3_1(1) = 1).
% 5.22/5.39  fof(interp, fi_functors, esk4_1(0) = 0 & esk4_1(1) = 1).
% 5.22/5.39  fof(interp, fi_functors, esk5_1(0) = 1 & esk5_1(1) = 1).
% 5.22/5.39  fof(interp, fi_functors, esk6_1(0) = 0 & esk6_1(1) = 0).
% 5.22/5.39  fof(interp, fi_functors, esk7_3(0, 0, 0) = 1 & esk7_3(0, 0, 1) = 1 &
% 5.22/5.39    esk7_3(0, 1, 0) = 0 &
% 5.22/5.39    esk7_3(0, 1, 1) = 1 &
% 5.22/5.39    esk7_3(1, 0, 0) = 0 &
% 5.22/5.39    esk7_3(1, 0, 1) = 0 &
% 5.22/5.39    esk7_3(1, 1, 0) = 0 &
% 5.22/5.39    esk7_3(1, 1, 1) = 1).
% 5.22/5.39  fof(interp, fi_functors, esk8_1(0) = 1 & esk8_1(1) = 1).
% 5.22/5.39  fof(interp, fi_functors, esk9_1(0) = 0 & esk9_1(1) = 1).
% 5.22/5.39  fof(interp, fi_predicates, ~p1(0) & p1(1)).
% 5.22/5.39  fof(interp, fi_predicates, r1(0, 0) & r1(0, 1) & ~r1(1, 0) & r1(1, 1)).
% 5.22/5.39  % SZS output end FiniteModel for theBenchmark.p
% 5.22/5.40  % 1 lemma(s) from E
% 5.22/5.40  %     cnf(cl, axiom, p1(esk11_0)).
% 5.22/5.40  % 43 pred(s)
% 5.22/5.40  % 12 func(s)
% 5.22/5.40  % 1 sort(s)
% 5.22/5.40  % 66 clause(s)
% 5.22/5.40  % Instantiating 1 (5031 ms)
% 5.22/5.40  % Solving (5031 ms)
% 5.22/5.40  % Instantiating 2 (5031 ms)
% 5.22/5.40  % Solving (5032 ms)
% 5.22/5.40  % 
% 5.22/5.40  % 1 model found (5032 ms)
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