%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : RNG101+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 : 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 13:59:17 EDT 2023

% Result   : Theorem 0.21s 0.60s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : RNG101+2 : 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
% 0.14/0.35  % Computer : n021.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Sun Aug 27 01:48:29 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.21/0.60  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.21/0.60  
% 0.21/0.60  % SZS status Theorem
% 0.21/0.60  
% 0.21/0.60  % SZS output start Proof
% 0.21/0.60  Take the following subset of the input axioms:
% 0.21/0.60    fof(m__, conjecture, ?[W0]: (aElement0(W0) & sdtasdt0(xc, W0)=xx)).
% 0.21/0.60    fof(m__1933, hypothesis, ?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=xx) & (aElementOf0(xx, slsdtgt0(xc)) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=xy) & (aElementOf0(xy, slsdtgt0(xc)) & aElement0(xz))))).
% 0.21/0.60  
% 0.21/0.60  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.60  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.60  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.60    fresh(y, y, x1...xn) = u
% 0.21/0.60    C => fresh(s, t, x1...xn) = v
% 0.21/0.60  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.60  variables of u and v.
% 0.21/0.60  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.60  input problem has no model of domain size 1).
% 0.21/0.60  
% 0.21/0.60  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.60  
% 0.21/0.61  Axiom 1 (m__1933_3): aElement0(w0_2) = true2.
% 0.21/0.61  Axiom 2 (m__1933): sdtasdt0(xc, w0_2) = xx.
% 0.21/0.61  
% 0.21/0.61  Goal 1 (m__): tuple(sdtasdt0(xc, X), aElement0(X)) = tuple(xx, true2).
% 0.21/0.61  The goal is true when:
% 0.21/0.61    X = w0_2
% 0.21/0.61  
% 0.21/0.61  Proof:
% 0.21/0.61    tuple(sdtasdt0(xc, w0_2), aElement0(w0_2))
% 0.21/0.61  = { by axiom 1 (m__1933_3) }
% 0.21/0.61    tuple(sdtasdt0(xc, w0_2), true2)
% 0.21/0.61  = { by axiom 2 (m__1933) }
% 0.21/0.61    tuple(xx, true2)
% 0.21/0.61  % SZS output end Proof
% 0.21/0.61  
% 0.21/0.61  RESULT: Theorem (the conjecture is true).
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