%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : LCL682+1.001 : TPTP v8.1.2. Released v4.0.0.
% 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 : n027.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 08:20:10 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
% 0.00/0.13  % Problem  : LCL682+1.001 : TPTP v8.1.2. Released v4.0.0.
% 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 : n027.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 : Thu Aug 24 19:23:47 EDT 2023
% 0.13/0.35  % 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:
% 0.20/0.40    fof(main, conjecture, ~?[X]: ~(~![Y]: (~r1(X, Y) | ~(~![X2]: (~r1(Y, X2) | p16(X2)) | (~![X2]: (~r1(Y, X2) | p12(X2)) | (~![X2]: (~r1(Y, X2) | p14(X2)) | ~![X2]: (~r1(Y, X2) | p12(X2)))))) | (![Y2]: (~r1(X, Y2) | ![X2]: (~r1(Y2, X2) | p15(X2))) | (![Y2]: (~r1(X, Y2) | ![X2]: (~r1(Y2, X2) | p13(X2))) | (![Y2]: (~r1(X, Y2) | ![X2]: (~r1(Y2, X2) | p12(X2))) | ![Y2]: (~r1(X, Y2) | ![X2]: (~r1(Y2, X2) | p11(X2)))))))).
% 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 (main_2): r1(x5, y2) = true.
% 0.20/0.40  Axiom 2 (main_6): r1(y2, x2) = true.
% 0.20/0.40  Axiom 3 (main_11): fresh7(X, X, Y) = true.
% 0.20/0.40  Axiom 4 (main_11): fresh8(X, X, Y, Z) = p12(Z).
% 0.20/0.40  Axiom 5 (main_11): fresh8(r1(x5, X), true, X, Y) = fresh7(r1(X, Y), true, Y).
% 0.20/0.40  
% 0.20/0.40  Goal 1 (main_12): p12(x2) = true.
% 0.20/0.40  Proof:
% 0.20/0.40    p12(x2)
% 0.20/0.40  = { by axiom 4 (main_11) R->L }
% 0.20/0.40    fresh8(true, true, y2, x2)
% 0.20/0.40  = { by axiom 1 (main_2) R->L }
% 0.20/0.40    fresh8(r1(x5, y2), true, y2, x2)
% 0.20/0.40  = { by axiom 5 (main_11) }
% 0.20/0.40    fresh7(r1(y2, x2), true, x2)
% 0.20/0.40  = { by axiom 2 (main_6) }
% 0.20/0.40    fresh7(true, true, x2)
% 0.20/0.40  = { by axiom 3 (main_11) }
% 0.20/0.40    true
% 0.20/0.40  % SZS output end Proof
% 0.20/0.40  
% 0.20/0.40  RESULT: Theorem (the conjecture is true).
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