%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : RNG102+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 : n025.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 13:59:18 EDT 2023

% Result   : Theorem 0.20s 0.62s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : RNG102+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
% 0.13/0.34  % Computer : n025.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Sun Aug 27 02:46:23 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.20/0.62  Command-line arguments: --no-flatten-goal
% 0.20/0.62  
% 0.20/0.62  % SZS status Theorem
% 0.20/0.62  
% 0.20/0.62  % SZS output start Proof
% 0.20/0.62  Take the following subset of the input axioms:
% 0.20/0.62    fof(m__, conjecture, ?[W0]: (aElement0(W0) & sdtasdt0(xc, W0)=xy)).
% 0.20/0.62    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))))).
% 0.20/0.62  
% 0.20/0.62  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.62  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.62  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.62    fresh(y, y, x1...xn) = u
% 0.20/0.62    C => fresh(s, t, x1...xn) = v
% 0.20/0.62  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.62  variables of u and v.
% 0.20/0.62  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.62  input problem has no model of domain size 1).
% 0.20/0.62  
% 0.20/0.62  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.62  
% 0.20/0.62  Axiom 1 (m__1933_4): aElement0(w0) = true2.
% 0.20/0.62  Axiom 2 (m__1933_1): sdtasdt0(xc, w0) = xy.
% 0.20/0.62  
% 0.20/0.62  Goal 1 (m__): tuple(sdtasdt0(xc, X), aElement0(X)) = tuple(xy, true2).
% 0.20/0.62  The goal is true when:
% 0.20/0.62    X = w0
% 0.20/0.62  
% 0.20/0.62  Proof:
% 0.20/0.62    tuple(sdtasdt0(xc, w0), aElement0(w0))
% 0.20/0.62  = { by axiom 1 (m__1933_4) }
% 0.20/0.62    tuple(sdtasdt0(xc, w0), true2)
% 0.20/0.62  = { by axiom 2 (m__1933_1) }
% 0.20/0.62    tuple(xy, true2)
% 0.20/0.62  % SZS output end Proof
% 0.20/0.62  
% 0.20/0.62  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------