%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : GEO498+1 : TPTP v8.1.2. Released v7.0.0.
% 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 : n001.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 30 23:29:06 EDT 2023

% Result   : Theorem 0.20s 0.75s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : GEO498+1 : TPTP v8.1.2. Released v7.0.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.18/0.34  % Computer : n001.cluster.edu
% 0.18/0.34  % Model    : x86_64 x86_64
% 0.18/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.18/0.34  % Memory   : 8042.1875MB
% 0.18/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.18/0.34  % CPULimit : 300
% 0.18/0.34  % WCLimit  : 300
% 0.18/0.34  % DateTime : Tue Aug 29 23:39:23 EDT 2023
% 0.18/0.34  % CPUTime  : 
% 0.20/0.75  Command-line arguments: --no-flatten-goal
% 0.20/0.75  
% 0.20/0.75  % SZS status Theorem
% 0.20/0.75  
% 0.20/0.75  % SZS output start Proof
% 0.20/0.75  Take the following subset of the input axioms:
% 0.20/0.75    fof(aSatz7_7, axiom, ![Xa, Xp]: s(Xa, s(Xa, Xp))=Xp).
% 0.20/0.75    fof(aSatz7_8, conjecture, ![Xq, Xr, Xa2, Xp2]: (s(Xa2, Xp2)!=Xr | (s(Xa2, Xq)!=Xr | Xp2=Xq))).
% 0.20/0.75  
% 0.20/0.75  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.75  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.75  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.75    fresh(y, y, x1...xn) = u
% 0.20/0.75    C => fresh(s, t, x1...xn) = v
% 0.20/0.75  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.75  variables of u and v.
% 0.20/0.75  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.75  input problem has no model of domain size 1).
% 0.20/0.75  
% 0.20/0.75  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.75  
% 0.20/0.75  Axiom 1 (aSatz7_8): s(xa, xp) = xr.
% 0.20/0.75  Axiom 2 (aSatz7_8_1): s(xa, xq) = xr.
% 0.20/0.75  Axiom 3 (aSatz7_7): s(X, s(X, Y)) = Y.
% 0.20/0.75  
% 0.20/0.75  Goal 1 (aSatz7_8_2): xp = xq.
% 0.20/0.75  Proof:
% 0.20/0.75    xp
% 0.20/0.75  = { by axiom 3 (aSatz7_7) R->L }
% 0.20/0.75    s(xa, s(xa, xp))
% 0.20/0.75  = { by axiom 1 (aSatz7_8) }
% 0.20/0.75    s(xa, xr)
% 0.20/0.75  = { by axiom 2 (aSatz7_8_1) R->L }
% 0.20/0.75    s(xa, s(xa, xq))
% 0.20/0.75  = { by axiom 3 (aSatz7_7) }
% 0.20/0.75    xq
% 0.20/0.75  % SZS output end Proof
% 0.20/0.75  
% 0.20/0.75  RESULT: Theorem (the conjecture is true).
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