%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUM617+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 11:57:18 EDT 2023

% Result   : Theorem 8.82s 1.56s
% Output   : Proof 8.82s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.14/0.14  % Problem  : NUM617+3 : TPTP v8.1.2. Released v4.0.0.
% 0.14/0.15  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.37  % Computer : n017.cluster.edu
% 0.15/0.37  % Model    : x86_64 x86_64
% 0.15/0.37  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37  % Memory   : 8042.1875MB
% 0.15/0.37  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37  % CPULimit : 300
% 0.15/0.37  % WCLimit  : 300
% 0.15/0.37  % DateTime : Fri Aug 25 11:33:25 EDT 2023
% 0.15/0.37  % CPUTime  : 
% 8.82/1.56  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 8.82/1.56  
% 8.82/1.56  % SZS status Theorem
% 8.82/1.56  
% 8.82/1.57  % SZS output start Proof
% 8.82/1.57  Take the following subset of the input axioms:
% 8.82/1.57    fof(m__, conjecture, aElementOf0(xx, szNzAzT0)).
% 8.82/1.57    fof(m__5106, hypothesis, ![W0]: (aElementOf0(W0, xQ) => aElementOf0(W0, szNzAzT0)) & aSubsetOf0(xQ, szNzAzT0)).
% 8.82/1.57    fof(m__5348, hypothesis, aElement0(xx) & (aElementOf0(xx, xQ) & (xx!=szmzizndt0(xQ) & aElementOf0(xx, xP)))).
% 8.82/1.57  
% 8.82/1.57  Now clausify the problem and encode Horn clauses using encoding 3 of
% 8.82/1.57  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 8.82/1.57  We repeatedly replace C & s=t => u=v by the two clauses:
% 8.82/1.57    fresh(y, y, x1...xn) = u
% 8.82/1.57    C => fresh(s, t, x1...xn) = v
% 8.82/1.57  where fresh is a fresh function symbol and x1..xn are the free
% 8.82/1.57  variables of u and v.
% 8.82/1.57  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 8.82/1.57  input problem has no model of domain size 1).
% 8.82/1.57  
% 8.82/1.57  The encoding turns the above axioms into the following unit equations and goals:
% 8.82/1.57  
% 8.82/1.57  Axiom 1 (m__5348_1): aElementOf0(xx, xQ) = true2.
% 8.82/1.57  Axiom 2 (m__5106_1): fresh24(X, X, Y) = true2.
% 8.82/1.57  Axiom 3 (m__5106_1): fresh24(aElementOf0(X, xQ), true2, X) = aElementOf0(X, szNzAzT0).
% 8.82/1.57  
% 8.82/1.57  Goal 1 (m__): aElementOf0(xx, szNzAzT0) = true2.
% 8.82/1.57  Proof:
% 8.82/1.57    aElementOf0(xx, szNzAzT0)
% 8.82/1.57  = { by axiom 3 (m__5106_1) R->L }
% 8.82/1.57    fresh24(aElementOf0(xx, xQ), true2, xx)
% 8.82/1.57  = { by axiom 1 (m__5348_1) }
% 8.82/1.57    fresh24(true2, true2, xx)
% 8.82/1.57  = { by axiom 2 (m__5106_1) }
% 8.82/1.57    true2
% 8.82/1.57  % SZS output end Proof
% 8.82/1.57  
% 8.82/1.57  RESULT: Theorem (the conjecture is true).
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