TSTP Solution File: NUM427+3 by SnakeForV-SAT---1.0

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%------------------------------------------------------------------------------
% File     : SnakeForV-SAT---1.0
% Problem  : NUM427+3 : TPTP v8.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_sat --cores 0 -t %d %s

% Computer : n002.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  : 300s
% DateTime : Wed Aug 31 18:05:04 EDT 2022

% Result   : Theorem 0.19s 0.51s
% Output   : Refutation 0.19s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    8
%            Number of leaves      :    7
% Syntax   : Number of formulae    :   22 (  12 unt;   0 def)
%            Number of atoms       :   38 (  15 equ)
%            Maximal formula atoms :    4 (   1 avg)
%            Number of connectives :   31 (  15   ~;  10   |;   5   &)
%                                         (   0 <=>;   1  =>;   0  <=;   0 <~>)
%            Maximal formula depth :    6 (   2 avg)
%            Maximal term depth    :    3 (   1 avg)
%            Number of predicates  :    5 (   3 usr;   1 prp; 0-3 aty)
%            Number of functors    :   10 (  10 usr;   6 con; 0-2 aty)
%            Number of variables   :    9 (   7   !;   2   ?)

% Comments : 
%------------------------------------------------------------------------------
fof(f335,plain,
    $false,
    inference(subsumption_resolution,[],[f334,f79]) ).

fof(f79,plain,
    aInteger0(xn),
    inference(cnf_transformation,[],[f23]) ).

fof(f23,axiom,
    ( aInteger0(xn)
    & sdtpldt0(xa,smndt0(xb)) = sdtasdt0(xq,xn) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',m__747) ).

fof(f334,plain,
    ~ aInteger0(xn),
    inference(resolution,[],[f333,f110]) ).

fof(f110,plain,
    ! [X0] :
      ( aInteger0(smndt0(X0))
      | ~ aInteger0(X0) ),
    inference(cnf_transformation,[],[f47]) ).

fof(f47,plain,
    ! [X0] :
      ( ~ aInteger0(X0)
      | aInteger0(smndt0(X0)) ),
    inference(ennf_transformation,[],[f4]) ).

fof(f4,axiom,
    ! [X0] :
      ( aInteger0(X0)
     => aInteger0(smndt0(X0)) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',mIntNeg) ).

fof(f333,plain,
    ~ aInteger0(smndt0(xn)),
    inference(trivial_inequality_removal,[],[f332]) ).

fof(f332,plain,
    ( ~ aInteger0(smndt0(xn))
    | sF3 != sF3 ),
    inference(superposition,[],[f128,f207]) ).

fof(f207,plain,
    sF4(smndt0(xn)) = sF3,
    inference(forward_demodulation,[],[f206,f125]) ).

fof(f125,plain,
    sF3 = sdtpldt0(xb,sF2),
    introduced(function_definition,[]) ).

fof(f206,plain,
    sF4(smndt0(xn)) = sdtpldt0(xb,sF2),
    inference(superposition,[],[f131,f124]) ).

fof(f124,plain,
    smndt0(xa) = sF2,
    introduced(function_definition,[]) ).

fof(f131,plain,
    sdtpldt0(xb,smndt0(xa)) = sF4(smndt0(xn)),
    inference(forward_demodulation,[],[f100,f127]) ).

fof(f127,plain,
    ! [X0] : sdtasdt0(xq,X0) = sF4(X0),
    introduced(function_definition,[]) ).

fof(f100,plain,
    sdtasdt0(xq,smndt0(xn)) = sdtpldt0(xb,smndt0(xa)),
    inference(cnf_transformation,[],[f24]) ).

fof(f24,axiom,
    sdtasdt0(xq,smndt0(xn)) = sdtpldt0(xb,smndt0(xa)),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',m__767) ).

fof(f128,plain,
    ! [X0] :
      ( sF3 != sF4(X0)
      | ~ aInteger0(X0) ),
    inference(definition_folding,[],[f118,f125,f124,f127]) ).

fof(f118,plain,
    ! [X0] :
      ( sdtasdt0(xq,X0) != sdtpldt0(xb,smndt0(xa))
      | ~ aInteger0(X0) ),
    inference(cnf_transformation,[],[f52]) ).

fof(f52,plain,
    ( ~ sdteqdtlpzmzozddtrp0(xb,xa,xq)
    & ~ aDivisorOf0(xq,sdtpldt0(xb,smndt0(xa)))
    & ! [X0] :
        ( sdtasdt0(xq,X0) != sdtpldt0(xb,smndt0(xa))
        | ~ aInteger0(X0) ) ),
    inference(ennf_transformation,[],[f26]) ).

fof(f26,negated_conjecture,
    ~ ( ? [X0] :
          ( aInteger0(X0)
          & sdtasdt0(xq,X0) = sdtpldt0(xb,smndt0(xa)) )
      | sdteqdtlpzmzozddtrp0(xb,xa,xq)
      | aDivisorOf0(xq,sdtpldt0(xb,smndt0(xa))) ),
    inference(negated_conjecture,[],[f25]) ).

fof(f25,conjecture,
    ( ? [X0] :
        ( aInteger0(X0)
        & sdtasdt0(xq,X0) = sdtpldt0(xb,smndt0(xa)) )
    | sdteqdtlpzmzozddtrp0(xb,xa,xq)
    | aDivisorOf0(xq,sdtpldt0(xb,smndt0(xa))) ),
    file('/export/starexec/sandbox/benchmark/theBenchmark.p',m__) ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem    : NUM427+3 : TPTP v8.1.0. Released v4.0.0.
% 0.03/0.13  % Command    : vampire --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_sat --cores 0 -t %d %s
% 0.12/0.34  % Computer : n002.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit   : 300
% 0.12/0.34  % WCLimit    : 300
% 0.12/0.34  % DateTime   : Tue Aug 30 06:43:13 EDT 2022
% 0.12/0.34  % CPUTime    : 
% 0.19/0.48  % (30909)ott+10_1:5_bd=off:tgt=full:i=99:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/99Mi)
% 0.19/0.49  % (30909)First to succeed.
% 0.19/0.49  % (30917)ott+3_1:1_gsp=on:lcm=predicate:i=138:si=on:rawr=on:rtra=on_0 on theBenchmark for (2999ds/138Mi)
% 0.19/0.51  % (30909)Refutation found. Thanks to Tanya!
% 0.19/0.51  % SZS status Theorem for theBenchmark
% 0.19/0.51  % SZS output start Proof for theBenchmark
% See solution above
% 0.19/0.51  % (30909)------------------------------
% 0.19/0.51  % (30909)Version: Vampire 4.7 (commit 807e37dd9 on 2022-08-23 09:55:27 +0200)
% 0.19/0.51  % (30909)Linked with Z3 4.8.13.0 f03d756e086f81f2596157241e0decfb1c982299 z3-4.8.4-5390-gf03d756e0
% 0.19/0.51  % (30909)Termination reason: Refutation
% 0.19/0.51  
% 0.19/0.51  % (30909)Memory used [KB]: 5628
% 0.19/0.51  % (30909)Time elapsed: 0.082 s
% 0.19/0.51  % (30909)Instructions burned: 10 (million)
% 0.19/0.51  % (30909)------------------------------
% 0.19/0.51  % (30909)------------------------------
% 0.19/0.51  % (30895)Success in time 0.159 s
%------------------------------------------------------------------------------