%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN318-1 : 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 : n031.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 03:34:03 EDT 2023

% Result   : Unsatisfiable 0.20s 0.38s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : SYN318-1 : TPTP v8.1.2. Released v1.2.0.
% 0.13/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n031.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit : 300
% 0.13/0.35  % WCLimit  : 300
% 0.13/0.35  % DateTime : Sat Aug 26 20:48:53 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.38  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.38  
% 0.20/0.38  % SZS status Unsatisfiable
% 0.20/0.39  
% 0.20/0.39  % SZS output start Proof
% 0.20/0.39  Take the following subset of the input axioms:
% 0.20/0.39    fof(clause1, negated_conjecture, ![X]: (~f(X) | (~f(b) | g(X)))).
% 0.20/0.39    fof(clause3, negated_conjecture, ![X2]: f(X2)).
% 0.20/0.39    fof(clause4, negated_conjecture, ~g(b)).
% 0.20/0.39  
% 0.20/0.39  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39    fresh(y, y, x1...xn) = u
% 0.20/0.39    C => fresh(s, t, x1...xn) = v
% 0.20/0.39  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39  variables of u and v.
% 0.20/0.39  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39  input problem has no model of domain size 1).
% 0.20/0.39  
% 0.20/0.39  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39  
% 0.20/0.39  Axiom 1 (clause3): f(X) = true.
% 0.20/0.39  Axiom 2 (clause1): fresh(X, X, Y) = g(Y).
% 0.20/0.39  Axiom 3 (clause1): fresh2(X, X, Y) = true.
% 0.20/0.39  Axiom 4 (clause1): fresh(f(b), true, X) = fresh2(f(X), true, X).
% 0.20/0.39  
% 0.20/0.39  Goal 1 (clause4): g(b) = true.
% 0.20/0.39  Proof:
% 0.20/0.39    g(b)
% 0.20/0.39  = { by axiom 2 (clause1) R->L }
% 0.20/0.39    fresh(true, true, b)
% 0.20/0.39  = { by axiom 1 (clause3) R->L }
% 0.20/0.39    fresh(f(b), true, b)
% 0.20/0.39  = { by axiom 4 (clause1) }
% 0.20/0.39    fresh2(f(b), true, b)
% 0.20/0.39  = { by axiom 1 (clause3) }
% 0.20/0.39    fresh2(true, true, b)
% 0.20/0.39  = { by axiom 3 (clause1) }
% 0.20/0.39    true
% 0.20/0.39  % SZS output end Proof
% 0.20/0.39  
% 0.20/0.39  RESULT: Unsatisfiable (the axioms are contradictory).
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