%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUM489+3 : 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 : n004.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 11:56:35 EDT 2023

% Result   : Theorem 0.21s 0.51s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : NUM489+3 : 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
% 0.14/0.34  % Computer : n004.cluster.edu
% 0.14/0.34  % Model    : x86_64 x86_64
% 0.14/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34  % Memory   : 8042.1875MB
% 0.14/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34  % CPULimit : 300
% 0.14/0.34  % WCLimit  : 300
% 0.14/0.34  % DateTime : Fri Aug 25 16:00:22 EDT 2023
% 0.14/0.34  % CPUTime  : 
% 0.21/0.51  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.51  
% 0.21/0.51  % SZS status Theorem
% 0.21/0.51  
% 0.21/0.51  % SZS output start Proof
% 0.21/0.51  Take the following subset of the input axioms:
% 0.21/0.51    fof(m__, conjecture, xn=sdtpldt0(xp, xr)).
% 0.21/0.51    fof(m__1883, hypothesis, aNaturalNumber0(xr) & (sdtpldt0(xp, xr)=xn & xr=sdtmndt0(xn, xp))).
% 0.21/0.51  
% 0.21/0.51  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.51  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.51  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.51    fresh(y, y, x1...xn) = u
% 0.21/0.51    C => fresh(s, t, x1...xn) = v
% 0.21/0.51  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.51  variables of u and v.
% 0.21/0.51  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.51  input problem has no model of domain size 1).
% 0.21/0.51  
% 0.21/0.51  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.51  
% 0.21/0.51  Axiom 1 (m__1883): sdtpldt0(xp, xr) = xn.
% 0.21/0.51  
% 0.21/0.51  Goal 1 (m__): xn = sdtpldt0(xp, xr).
% 0.21/0.51  Proof:
% 0.21/0.51    xn
% 0.21/0.51  = { by axiom 1 (m__1883) R->L }
% 0.21/0.51    sdtpldt0(xp, xr)
% 0.21/0.51  % SZS output end Proof
% 0.21/0.51  
% 0.21/0.51  RESULT: Theorem (the conjecture is true).
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