%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUN069+1 : 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 : n004.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 0.49s
% 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  : NUN069+1 : 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
% 0.13/0.35  % Computer : n004.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 : Sun Aug 27 09:55:07 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.49  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.49  
% 0.20/0.49  % SZS status Theorem
% 0.20/0.49  
% 0.20/0.49  % SZS output start Proof
% 0.20/0.49  Take the following subset of the input axioms:
% 0.20/0.49    fof(axiom_1a, axiom, ![X1, X8]: ?[Y4]: (?[Y5]: (id(Y5, Y4) & ?[Y15]: (r2(X8, Y15) & r3(X1, Y15, Y5))) & ?[Y7]: (r2(Y7, Y4) & r3(X1, X8, Y7)))).
% 0.20/0.49    fof(axiom_4a, axiom, ![X4]: ?[Y9]: (id(Y9, X4) & ?[Y16]: (r1(Y16) & r3(X4, Y16, Y9)))).
% 0.20/0.49    fof(axiom_5, axiom, ![X20]: id(X20, X20)).
% 0.20/0.49    fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~id(Y20, Y10) | ~r1(Y20)) | ~r2(X7, Y10))).
% 0.20/0.49    fof(oneidone, conjecture, ?[Y1]: (id(Y1, Y1) & ?[Y2]: (r1(Y2) & r2(Y2, Y1)))).
% 0.20/0.49  
% 0.20/0.49  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.49  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.49  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.49    fresh(y, y, x1...xn) = u
% 0.20/0.49    C => fresh(s, t, x1...xn) = v
% 0.20/0.49  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.49  variables of u and v.
% 0.20/0.49  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.49  input problem has no model of domain size 1).
% 0.20/0.49  
% 0.20/0.49  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.49  
% 0.20/0.49  Axiom 1 (axiom_5): id(X, X) = true2.
% 0.20/0.49  Axiom 2 (axiom_4a_1): r1(y16(X)) = true2.
% 0.20/0.49  Axiom 3 (axiom_1a_1): r2(X, y15(Y, X)) = true2.
% 0.20/0.49  
% 0.20/0.49  Goal 1 (oneidone): tuple(id(X, X), r1(Y), r2(Y, X)) = tuple(true2, true2, true2).
% 0.20/0.49  The goal is true when:
% 0.20/0.49    X = y15(Y, y16(X))
% 0.20/0.49    Y = y16(X)
% 0.20/0.49  
% 0.20/0.49  Proof:
% 0.20/0.49    tuple(id(y15(Y, y16(X)), y15(Y, y16(X))), r1(y16(X)), r2(y16(X), y15(Y, y16(X))))
% 0.20/0.49  = { by axiom 1 (axiom_5) }
% 0.20/0.49    tuple(true2, r1(y16(X)), r2(y16(X), y15(Y, y16(X))))
% 0.20/0.49  = { by axiom 3 (axiom_1a_1) }
% 0.20/0.49    tuple(true2, r1(y16(X)), true2)
% 0.20/0.49  = { by axiom 2 (axiom_4a_1) }
% 0.20/0.49    tuple(true2, true2, true2)
% 0.20/0.49  % SZS output end Proof
% 0.20/0.49  
% 0.20/0.49  RESULT: Theorem (the conjecture is true).
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