%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN284-1 : TPTP v8.1.2. Released v1.1.0.
% 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 : n002.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 03:33:55 EDT 2023

% Result   : Unsatisfiable 29.40s 4.25s
% Output   : Proof 29.40s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SYN284-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34  % Computer : n002.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 300
% 0.12/0.34  % DateTime : Sat Aug 26 17:52:18 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 29.40/4.25  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 29.40/4.25  
% 29.40/4.25  % SZS status Unsatisfiable
% 29.40/4.25  
% 29.40/4.25  % SZS output start Proof
% 29.40/4.25  Take the following subset of the input axioms:
% 29.40/4.25    fof(axiom_14, axiom, ![X]: p0(b, X)).
% 29.40/4.25    fof(axiom_26, axiom, n0(d, c)).
% 29.40/4.25    fof(prove_this, negated_conjecture, ~q1(a, c, a)).
% 29.40/4.25    fof(rule_092, axiom, ![J, C, B, A2]: (q1(J, A2, J) | (~n0(B, A2) | ~p0(C, J)))).
% 29.40/4.25  
% 29.40/4.25  Now clausify the problem and encode Horn clauses using encoding 3 of
% 29.40/4.25  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 29.40/4.25  We repeatedly replace C & s=t => u=v by the two clauses:
% 29.40/4.25    fresh(y, y, x1...xn) = u
% 29.40/4.25    C => fresh(s, t, x1...xn) = v
% 29.40/4.25  where fresh is a fresh function symbol and x1..xn are the free
% 29.40/4.25  variables of u and v.
% 29.40/4.25  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 29.40/4.25  input problem has no model of domain size 1).
% 29.40/4.25  
% 29.40/4.25  The encoding turns the above axioms into the following unit equations and goals:
% 29.40/4.25  
% 29.40/4.25  Axiom 1 (axiom_14): p0(b, X) = true.
% 29.40/4.25  Axiom 2 (axiom_26): n0(d, c) = true.
% 29.40/4.25  Axiom 3 (rule_092): fresh317(X, X, Y, Z) = true.
% 29.40/4.25  Axiom 4 (rule_092): fresh318(X, X, Y, Z, W) = q1(Y, Z, Y).
% 29.40/4.26  Axiom 5 (rule_092): fresh318(p0(X, Y), true, Y, Z, W) = fresh317(n0(W, Z), true, Y, Z).
% 29.40/4.26  
% 29.40/4.26  Goal 1 (prove_this): q1(a, c, a) = true.
% 29.40/4.26  Proof:
% 29.40/4.26    q1(a, c, a)
% 29.40/4.26  = { by axiom 4 (rule_092) R->L }
% 29.40/4.26    fresh318(true, true, a, c, d)
% 29.40/4.26  = { by axiom 1 (axiom_14) R->L }
% 29.40/4.26    fresh318(p0(b, a), true, a, c, d)
% 29.40/4.26  = { by axiom 5 (rule_092) }
% 29.40/4.26    fresh317(n0(d, c), true, a, c)
% 29.40/4.26  = { by axiom 2 (axiom_26) }
% 29.40/4.26    fresh317(true, true, a, c)
% 29.40/4.26  = { by axiom 3 (rule_092) }
% 29.40/4.26    true
% 29.40/4.26  % SZS output end Proof
% 29.40/4.26  
% 29.40/4.26  RESULT: Unsatisfiable (the axioms are contradictory).
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