%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN063-2 : TPTP v8.1.2. Released v1.2.0.
% 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 : n004.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 : Fri Sep  1 03:32:58 EDT 2023

% Result   : Unsatisfiable 0.07s 0.32s
% Output   : Proof 0.07s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.08  % Problem  : SYN063-2 : TPTP v8.1.2. Released v1.2.0.
% 0.00/0.09  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.07/0.29  % Computer : n004.cluster.edu
% 0.07/0.29  % Model    : x86_64 x86_64
% 0.07/0.29  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.07/0.29  % Memory   : 8042.1875MB
% 0.07/0.29  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.07/0.29  % CPULimit : 300
% 0.07/0.29  % WCLimit  : 300
% 0.07/0.29  % DateTime : Sat Aug 26 20:29:52 EDT 2023
% 0.07/0.29  % CPUTime  : 
% 0.07/0.32  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.07/0.32  
% 0.07/0.32  % SZS status Unsatisfiable
% 0.07/0.32  
% 0.07/0.32  % SZS output start Proof
% 0.07/0.32  Take the following subset of the input axioms:
% 0.07/0.32    fof(pel33_1, negated_conjecture, big_p(c) | ~big_p(a)).
% 0.07/0.32    fof(pel33_2, negated_conjecture, big_p(a)).
% 0.07/0.32    fof(pel33_3, negated_conjecture, ~big_p(c)).
% 0.07/0.32  
% 0.07/0.32  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.07/0.32  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.07/0.32  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.07/0.32    fresh(y, y, x1...xn) = u
% 0.07/0.32    C => fresh(s, t, x1...xn) = v
% 0.07/0.32  where fresh is a fresh function symbol and x1..xn are the free
% 0.07/0.32  variables of u and v.
% 0.07/0.32  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.07/0.32  input problem has no model of domain size 1).
% 0.07/0.32  
% 0.07/0.32  The encoding turns the above axioms into the following unit equations and goals:
% 0.07/0.32  
% 0.07/0.32  Axiom 1 (pel33_2): big_p(a) = true.
% 0.07/0.32  Axiom 2 (pel33_1): fresh(X, X) = true.
% 0.07/0.32  Axiom 3 (pel33_1): fresh(big_p(a), true) = big_p(c).
% 0.07/0.32  
% 0.07/0.32  Goal 1 (pel33_3): big_p(c) = true.
% 0.07/0.32  Proof:
% 0.07/0.32    big_p(c)
% 0.07/0.32  = { by axiom 3 (pel33_1) R->L }
% 0.07/0.32    fresh(big_p(a), true)
% 0.07/0.32  = { by axiom 1 (pel33_2) }
% 0.07/0.32    fresh(true, true)
% 0.07/0.32  = { by axiom 2 (pel33_1) }
% 0.07/0.32    true
% 0.07/0.32  % SZS output end Proof
% 0.07/0.32  
% 0.07/0.32  RESULT: Unsatisfiable (the axioms are contradictory).
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