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% File     : Twee---2.4.2
% Problem  : SYN277-1 : 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 : n011.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 03:33:53 EDT 2023

% Result   : Unsatisfiable 13.22s 2.06s
% Output   : Proof 13.22s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.13  % Problem  : SYN277-1 : TPTP v8.1.2. Released v1.1.0.
% 0.12/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.35  % Computer : n011.cluster.edu
% 0.15/0.35  % Model    : x86_64 x86_64
% 0.15/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.35  % Memory   : 8042.1875MB
% 0.15/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.15/0.35  % CPULimit : 300
% 0.15/0.35  % WCLimit  : 300
% 0.15/0.35  % DateTime : Sat Aug 26 19:12:54 EDT 2023
% 0.15/0.35  % CPUTime  : 
% 13.22/2.06  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 13.22/2.06  
% 13.22/2.06  % SZS status Unsatisfiable
% 13.22/2.06  
% 13.22/2.07  % SZS output start Proof
% 13.22/2.07  Take the following subset of the input axioms:
% 13.22/2.07    fof(axiom_1, axiom, s0(d)).
% 13.22/2.07    fof(axiom_28, axiom, k0(e)).
% 13.22/2.07    fof(prove_this, negated_conjecture, ![X]: ~q1(X, X, d)).
% 13.22/2.07    fof(rule_117, axiom, q1(d, d, d) | (~k0(e) | ~s0(d))).
% 13.22/2.07  
% 13.22/2.07  Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.22/2.07  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.22/2.07  We repeatedly replace C & s=t => u=v by the two clauses:
% 13.22/2.07    fresh(y, y, x1...xn) = u
% 13.22/2.07    C => fresh(s, t, x1...xn) = v
% 13.22/2.07  where fresh is a fresh function symbol and x1..xn are the free
% 13.22/2.07  variables of u and v.
% 13.22/2.07  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.22/2.07  input problem has no model of domain size 1).
% 13.22/2.07  
% 13.22/2.07  The encoding turns the above axioms into the following unit equations and goals:
% 13.22/2.07  
% 13.22/2.07  Axiom 1 (axiom_1): s0(d) = true2.
% 13.22/2.07  Axiom 2 (axiom_28): k0(e) = true2.
% 13.22/2.07  Axiom 3 (rule_117): fresh285(X, X) = true2.
% 13.22/2.07  Axiom 4 (rule_117): fresh286(X, X) = q1(d, d, d).
% 13.22/2.07  Axiom 5 (rule_117): fresh286(k0(e), true2) = fresh285(s0(d), true2).
% 13.22/2.07  
% 13.22/2.07  Goal 1 (prove_this): q1(X, X, d) = true2.
% 13.22/2.07  The goal is true when:
% 13.22/2.07    X = d
% 13.22/2.07  
% 13.22/2.07  Proof:
% 13.22/2.07    q1(d, d, d)
% 13.22/2.07  = { by axiom 4 (rule_117) R->L }
% 13.22/2.07    fresh286(true2, true2)
% 13.22/2.07  = { by axiom 2 (axiom_28) R->L }
% 13.22/2.07    fresh286(k0(e), true2)
% 13.22/2.07  = { by axiom 5 (rule_117) }
% 13.22/2.07    fresh285(s0(d), true2)
% 13.22/2.07  = { by axiom 1 (axiom_1) }
% 13.22/2.07    fresh285(true2, true2)
% 13.22/2.07  = { by axiom 3 (rule_117) }
% 13.22/2.07    true2
% 13.22/2.07  % SZS output end Proof
% 13.22/2.07  
% 13.22/2.07  RESULT: Unsatisfiable (the axioms are contradictory).
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