%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUN067+2 : TPTP v8.1.2. Released v7.3.0.
% 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 : n019.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 : Thu Aug 31 12:51:49 EDT 2023

% Result   : Theorem 0.20s 0.41s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : NUN067+2 : TPTP v8.1.2. Released v7.3.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n019.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % 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 : Sun Aug 27 09:16:28 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.41  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.41  
% 0.20/0.41  % SZS status Theorem
% 0.20/0.41  
% 0.20/0.41  % SZS output start Proof
% 0.20/0.41  Take the following subset of the input axioms:
% 0.20/0.41    fof(axiom_1, axiom, ?[Y24]: ![X19]: ((~r1(X19) & X19!=Y24) | (r1(X19) & X19=Y24))).
% 0.20/0.41    fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.41    fof(nonzerosexist, conjecture, ?[Y1]: ![Y2]: (~r1(Y2) | Y1!=Y2)).
% 0.20/0.41  
% 0.20/0.41  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.41  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.41  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.41    fresh(y, y, x1...xn) = u
% 0.20/0.41    C => fresh(s, t, x1...xn) = v
% 0.20/0.41  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.41  variables of u and v.
% 0.20/0.41  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.41  input problem has no model of domain size 1).
% 0.20/0.41  
% 0.20/0.41  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.41  
% 0.20/0.41  Axiom 1 (nonzerosexist): X = y2(X).
% 0.20/0.41  Axiom 2 (nonzerosexist_1): r1(y2(X)) = true2.
% 0.20/0.41  Axiom 3 (axiom_1_1): fresh10(X, X, Y) = y24.
% 0.20/0.41  Axiom 4 (axiom_1_1): fresh10(r1(X), true2, X) = X.
% 0.20/0.41  
% 0.20/0.41  Lemma 5: r1(X) = true2.
% 0.20/0.41  Proof:
% 0.20/0.41    r1(X)
% 0.20/0.41  = { by axiom 1 (nonzerosexist) }
% 0.20/0.41    r1(y2(X))
% 0.20/0.41  = { by axiom 2 (nonzerosexist_1) }
% 0.20/0.41    true2
% 0.20/0.41  
% 0.20/0.41  Goal 1 (true_equals_false): true = false.
% 0.20/0.41  Proof:
% 0.20/0.41    true
% 0.20/0.41  = { by axiom 4 (axiom_1_1) R->L }
% 0.20/0.41    fresh10(r1(true), true2, true)
% 0.20/0.41  = { by lemma 5 }
% 0.20/0.41    fresh10(true2, true2, true)
% 0.20/0.41  = { by axiom 3 (axiom_1_1) }
% 0.20/0.41    y24
% 0.20/0.41  = { by axiom 3 (axiom_1_1) R->L }
% 0.20/0.41    fresh10(true2, true2, false)
% 0.20/0.41  = { by lemma 5 R->L }
% 0.20/0.41    fresh10(r1(false), true2, false)
% 0.20/0.41  = { by axiom 4 (axiom_1_1) }
% 0.20/0.41    false
% 0.20/0.42  % SZS output end Proof
% 0.20/0.42  
% 0.20/0.42  RESULT: Theorem (the conjecture is true).
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