%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : RNG074+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 : n017.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 13:59:09 EDT 2023

% Result   : Theorem 0.20s 0.55s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.13  % Problem  : RNG074+2 : TPTP v8.1.2. Released v4.0.0.
% 0.08/0.14  % 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 : n017.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 : Sun Aug 27 01:00:56 EDT 2023
% 0.13/0.36  % CPUTime  : 
% 0.20/0.55  Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.55  
% 0.20/0.55  % SZS status Theorem
% 0.20/0.55  
% 0.20/0.55  % SZS output start Proof
% 0.20/0.55  Take the following subset of the input axioms:
% 0.20/0.55    fof(m__2590, hypothesis, ~sdtlseqdt0(sdtpldt0(xP, xP), sdtpldt0(xR, xS))).
% 0.20/0.55    fof(m__2610, hypothesis, sdtlseqdt0(sdtpldt0(xR, xS), sdtpldt0(xP, xP))).
% 0.20/0.55    fof(m__2679, hypothesis, sdtpldt0(xR, xS)=sdtpldt0(xP, xP)).
% 0.20/0.55  
% 0.20/0.55  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.55  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.55  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.55    fresh(y, y, x1...xn) = u
% 0.20/0.55    C => fresh(s, t, x1...xn) = v
% 0.20/0.55  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.55  variables of u and v.
% 0.20/0.55  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.55  input problem has no model of domain size 1).
% 0.20/0.55  
% 0.20/0.55  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.55  
% 0.20/0.55  Axiom 1 (m__2679): sdtpldt0(xR, xS) = sdtpldt0(xP, xP).
% 0.20/0.55  Axiom 2 (m__2610): sdtlseqdt0(sdtpldt0(xR, xS), sdtpldt0(xP, xP)) = true2.
% 0.20/0.55  
% 0.20/0.55  Goal 1 (m__2590): sdtlseqdt0(sdtpldt0(xP, xP), sdtpldt0(xR, xS)) = true2.
% 0.20/0.55  Proof:
% 0.20/0.55    sdtlseqdt0(sdtpldt0(xP, xP), sdtpldt0(xR, xS))
% 0.20/0.55  = { by axiom 1 (m__2679) R->L }
% 0.20/0.55    sdtlseqdt0(sdtpldt0(xR, xS), sdtpldt0(xR, xS))
% 0.20/0.55  = { by axiom 1 (m__2679) }
% 0.20/0.55    sdtlseqdt0(sdtpldt0(xR, xS), sdtpldt0(xP, xP))
% 0.20/0.55  = { by axiom 2 (m__2610) }
% 0.20/0.55    true2
% 0.20/0.55  % SZS output end Proof
% 0.20/0.55  
% 0.20/0.55  RESULT: Theorem (the conjecture is true).
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