%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : RNG087+2 : 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 : n009.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 13:59:13 EDT 2023

% Result   : Theorem 0.19s 0.53s
% Output   : Proof 0.19s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12  % Problem  : RNG087+2 : TPTP v8.1.2. Released v4.0.0.
% 0.06/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.33  % Computer : n009.cluster.edu
% 0.13/0.33  % Model    : x86_64 x86_64
% 0.13/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33  % Memory   : 8042.1875MB
% 0.13/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33  % CPULimit : 300
% 0.13/0.33  % WCLimit  : 300
% 0.13/0.33  % DateTime : Sun Aug 27 02:13:20 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.19/0.53  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.53  
% 0.19/0.53  % SZS status Theorem
% 0.19/0.53  
% 0.19/0.53  % SZS output start Proof
% 0.19/0.53  Take the following subset of the input axioms:
% 0.19/0.53    fof(m__, conjecture, ?[W0, W1]: (aElementOf0(W0, xI) & (aElementOf0(W1, xJ) & xy=sdtpldt0(W0, W1)))).
% 0.19/0.53    fof(m__901, hypothesis, ?[W0_2, W1_2]: (aElementOf0(W0_2, xI) & (aElementOf0(W1_2, xJ) & sdtpldt0(W0_2, W1_2)=xx)) & (aElementOf0(xx, sdtpldt1(xI, xJ)) & (?[W0_2, W1_2]: (aElementOf0(W0_2, xI) & (aElementOf0(W1_2, xJ) & sdtpldt0(W0_2, W1_2)=xy)) & (aElementOf0(xy, sdtpldt1(xI, xJ)) & aElement0(xz))))).
% 0.19/0.53  
% 0.19/0.53  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.53  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.53  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.53    fresh(y, y, x1...xn) = u
% 0.19/0.53    C => fresh(s, t, x1...xn) = v
% 0.19/0.53  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.53  variables of u and v.
% 0.19/0.53  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.53  input problem has no model of domain size 1).
% 0.19/0.53  
% 0.19/0.53  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.53  
% 0.19/0.53  Axiom 1 (m__901_7): aElementOf0(w1, xJ) = true2.
% 0.19/0.54  Axiom 2 (m__901_8): aElementOf0(w0, xI) = true2.
% 0.19/0.54  Axiom 3 (m__901_1): sdtpldt0(w0, w1) = xy.
% 0.19/0.54  
% 0.19/0.54  Goal 1 (m__): tuple(xy, aElementOf0(X, xI), aElementOf0(Y, xJ)) = tuple(sdtpldt0(X, Y), true2, true2).
% 0.19/0.54  The goal is true when:
% 0.19/0.54    X = w0
% 0.19/0.54    Y = w1
% 0.19/0.54  
% 0.19/0.54  Proof:
% 0.19/0.54    tuple(xy, aElementOf0(w0, xI), aElementOf0(w1, xJ))
% 0.19/0.54  = { by axiom 2 (m__901_8) }
% 0.19/0.54    tuple(xy, true2, aElementOf0(w1, xJ))
% 0.19/0.54  = { by axiom 1 (m__901_7) }
% 0.19/0.54    tuple(xy, true2, true2)
% 0.19/0.54  = { by axiom 3 (m__901_1) R->L }
% 0.19/0.54    tuple(sdtpldt0(w0, w1), true2, true2)
% 0.19/0.54  % SZS output end Proof
% 0.19/0.54  
% 0.19/0.54  RESULT: Theorem (the conjecture is true).
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