%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN050-1 : TPTP v8.1.2. Released v1.0.0.
% 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 : n003.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 03:32:52 EDT 2023

% Result   : Unsatisfiable 0.20s 0.39s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SYN050-1 : TPTP v8.1.2. Released v1.0.0.
% 0.12/0.13  % 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 : n003.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 17:16:39 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.39  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.39  
% 0.20/0.39  % 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(clause_1, negated_conjecture, ![Y, Z]: (~big_p(Y) | (~big_q(Z) | big_r(f(Y, Z))))).
% 0.20/0.39    fof(clause_3, negated_conjecture, big_p(a)).
% 0.20/0.39    fof(clause_4, negated_conjecture, big_q(b)).
% 0.20/0.39    fof(clause_5, negated_conjecture, ![W]: ~big_r(W)).
% 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  
% 0.20/0.40  Axiom 1 (clause_3): big_p(a) = true2.
% 0.20/0.40  Axiom 2 (clause_4): big_q(b) = true2.
% 0.20/0.40  Axiom 3 (clause_1): fresh4(X, X, Y, Z) = true2.
% 0.20/0.40  Axiom 4 (clause_1): fresh3(X, X, Y, Z) = big_r(f(Y, Z)).
% 0.20/0.40  Axiom 5 (clause_1): fresh3(big_q(X), true2, Y, X) = fresh4(big_p(Y), true2, Y, X).
% 0.20/0.40  
% 0.20/0.40  Goal 1 (clause_5): big_r(X) = true2.
% 0.20/0.40  The goal is true when:
% 0.20/0.40    X = f(a, b)
% 0.20/0.40  
% 0.20/0.40  Proof:
% 0.20/0.40    big_r(f(a, b))
% 0.20/0.40  = { by axiom 4 (clause_1) R->L }
% 0.20/0.40    fresh3(true2, true2, a, b)
% 0.20/0.40  = { by axiom 2 (clause_4) R->L }
% 0.20/0.40    fresh3(big_q(b), true2, a, b)
% 0.20/0.40  = { by axiom 5 (clause_1) }
% 0.20/0.40    fresh4(big_p(a), true2, a, b)
% 0.20/0.40  = { by axiom 1 (clause_3) }
% 0.20/0.40    fresh4(true2, true2, a, b)
% 0.20/0.40  = { by axiom 3 (clause_1) }
% 0.20/0.40    true2
% 0.20/0.40  % SZS output end Proof
% 0.20/0.40  
% 0.20/0.40  RESULT: Unsatisfiable (the axioms are contradictory).
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