%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUN069+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 : n021.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 12:51:50 EDT 2023

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

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : NUN069+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
% 0.18/0.34  % Computer : n021.cluster.edu
% 0.18/0.34  % Model    : x86_64 x86_64
% 0.18/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.18/0.34  % Memory   : 8042.1875MB
% 0.18/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.18/0.34  % CPULimit : 300
% 0.18/0.34  % WCLimit  : 300
% 0.18/0.34  % DateTime : Sun Aug 27 09:17:57 EDT 2023
% 0.18/0.34  % CPUTime  : 
% 0.20/0.39  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.39  
% 0.20/0.39  % SZS status Theorem
% 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(axiom_1a, axiom, ![X1, X8]: ?[Y4]: (?[Y5]: (?[Y15]: (r2(X8, Y15) & r3(X1, Y15, Y5)) & Y5=Y4) & ?[Y7]: (r2(Y7, Y4) & r3(X1, X8, Y7)))).
% 0.20/0.39    fof(axiom_4a, axiom, ![X4]: ?[Y9]: (?[Y16]: (r1(Y16) & r3(X4, Y16, Y9)) & Y9=X4)).
% 0.20/0.39    fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~r1(Y20) | Y20!=Y10) | ~r2(X7, Y10))).
% 0.20/0.40    fof(oneeqone, conjecture, ?[Y1]: (Y1=Y1 & ?[Y2]: (r1(Y2) & r2(Y2, Y1)))).
% 0.20/0.40  
% 0.20/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.40    fresh(y, y, x1...xn) = u
% 0.20/0.40    C => fresh(s, t, x1...xn) = v
% 0.20/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.40  variables of u and v.
% 0.20/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.40  input problem has no model of domain size 1).
% 0.20/0.40  
% 0.20/0.40  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.40  
% 0.20/0.40  Axiom 1 (axiom_1a_1): r2(X, y15(Y, X)) = true2.
% 0.20/0.40  Axiom 2 (axiom_4a_1): r1(y16(X)) = true2.
% 0.20/0.40  
% 0.20/0.40  Goal 1 (oneeqone): tuple(r1(X), r2(X, Y)) = tuple(true2, true2).
% 0.20/0.40  The goal is true when:
% 0.20/0.40    X = y16(X)
% 0.20/0.40    Y = y15(Y, y16(X))
% 0.20/0.40  
% 0.20/0.40  Proof:
% 0.20/0.40    tuple(r1(y16(X)), r2(y16(X), y15(Y, y16(X))))
% 0.20/0.40  = { by axiom 1 (axiom_1a_1) }
% 0.20/0.40    tuple(r1(y16(X)), true2)
% 0.20/0.40  = { by axiom 2 (axiom_4a_1) }
% 0.20/0.40    tuple(true2, true2)
% 0.20/0.40  % SZS output end Proof
% 0.20/0.40  
% 0.20/0.40  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------