%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUM625+3 : TPTP v8.1.2. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof

% Computer : n017.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 : Thu Aug 31 11:57:21 EDT 2023

% Result   : Theorem 13.34s 2.24s
% Output   : Proof 13.34s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.10  % Problem  : NUM625+3 : TPTP v8.1.2. Released v4.0.0.
% 0.08/0.11  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.10/0.31  % Computer : n017.cluster.edu
% 0.10/0.31  % Model    : x86_64 x86_64
% 0.10/0.31  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.31  % Memory   : 8042.1875MB
% 0.10/0.31  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.10/0.31  % CPULimit : 300
% 0.10/0.31  % WCLimit  : 300
% 0.10/0.31  % DateTime : Fri Aug 25 13:29:25 EDT 2023
% 0.10/0.31  % CPUTime  : 
% 13.34/2.24  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 13.34/2.24  
% 13.34/2.24  % SZS status Theorem
% 13.34/2.24  
% 13.34/2.25  % SZS output start Proof
% 13.34/2.25  Take the following subset of the input axioms:
% 13.34/2.25    fof(m__5147, hypothesis, aElementOf0(xp, xQ) & (![W0]: (aElementOf0(W0, xQ) => sdtlseqdt0(xp, W0)) & xp=szmzizndt0(xQ))).
% 13.34/2.25    fof(m__5348, hypothesis, aElement0(xx) & (aElementOf0(xx, xQ) & (xx!=szmzizndt0(xQ) & aElementOf0(xx, xP)))).
% 13.34/2.25    fof(m__5496, hypothesis, xp=xx).
% 13.34/2.25  
% 13.34/2.25  Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.34/2.25  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.34/2.25  We repeatedly replace C & s=t => u=v by the two clauses:
% 13.34/2.25    fresh(y, y, x1...xn) = u
% 13.34/2.25    C => fresh(s, t, x1...xn) = v
% 13.34/2.25  where fresh is a fresh function symbol and x1..xn are the free
% 13.34/2.25  variables of u and v.
% 13.34/2.25  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.34/2.25  input problem has no model of domain size 1).
% 13.34/2.25  
% 13.34/2.25  The encoding turns the above axioms into the following unit equations and goals:
% 13.34/2.25  
% 13.34/2.25  Axiom 1 (m__5496): xp = xx.
% 13.34/2.25  Axiom 2 (m__5147): xp = szmzizndt0(xQ).
% 13.34/2.25  
% 13.34/2.25  Goal 1 (m__5348_3): xx = szmzizndt0(xQ).
% 13.34/2.25  Proof:
% 13.34/2.25    xx
% 13.34/2.25  = { by axiom 1 (m__5496) R->L }
% 13.34/2.25    xp
% 13.34/2.25  = { by axiom 2 (m__5147) }
% 13.34/2.25    szmzizndt0(xQ)
% 13.34/2.25  % SZS output end Proof
% 13.34/2.25  
% 13.34/2.25  RESULT: Theorem (the conjecture is true).
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