%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUM632+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 : n029.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 11:57:24 EDT 2023

% Result   : Theorem 11.67s 1.88s
% Output   : Proof 11.67s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : NUM632+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
% 0.12/0.33  % Computer : n029.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit : 300
% 0.12/0.33  % WCLimit  : 300
% 0.12/0.33  % DateTime : Fri Aug 25 10:27:09 EDT 2023
% 0.12/0.33  % CPUTime  : 
% 11.67/1.88  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 11.67/1.88  
% 11.67/1.88  % SZS status Theorem
% 11.67/1.88  
% 11.67/1.89  % SZS output start Proof
% 11.67/1.89  Take the following subset of the input axioms:
% 11.67/1.89    fof(m__, conjecture, sdtlpdtrp0(xc, xQ)=szDzizrdt0(xd)).
% 11.67/1.89    fof(m__5309, hypothesis, aElementOf0(xn, szDzozmdt0(xd)) & (sdtlpdtrp0(xd, xn)=szDzizrdt0(xd) & (aElementOf0(xn, sdtlbdtrb0(xd, szDzizrdt0(xd))) & (aElementOf0(xn, szNzAzT0) & sdtlpdtrp0(xe, xn)=xp)))).
% 11.67/1.89    fof(m__5568, hypothesis, sdtlpdtrp0(xc, xQ)=sdtlpdtrp0(xd, xn)).
% 11.67/1.89  
% 11.67/1.89  Now clausify the problem and encode Horn clauses using encoding 3 of
% 11.67/1.89  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 11.67/1.89  We repeatedly replace C & s=t => u=v by the two clauses:
% 11.67/1.89    fresh(y, y, x1...xn) = u
% 11.67/1.89    C => fresh(s, t, x1...xn) = v
% 11.67/1.89  where fresh is a fresh function symbol and x1..xn are the free
% 11.67/1.89  variables of u and v.
% 11.67/1.89  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 11.67/1.89  input problem has no model of domain size 1).
% 11.67/1.89  
% 11.67/1.89  The encoding turns the above axioms into the following unit equations and goals:
% 11.67/1.89  
% 11.67/1.89  Axiom 1 (m__5309_1): sdtlpdtrp0(xd, xn) = szDzizrdt0(xd).
% 11.67/1.89  Axiom 2 (m__5568): sdtlpdtrp0(xc, xQ) = sdtlpdtrp0(xd, xn).
% 11.67/1.89  
% 11.67/1.89  Goal 1 (m__): sdtlpdtrp0(xc, xQ) = szDzizrdt0(xd).
% 11.67/1.89  Proof:
% 11.67/1.89    sdtlpdtrp0(xc, xQ)
% 11.67/1.89  = { by axiom 2 (m__5568) }
% 11.67/1.89    sdtlpdtrp0(xd, xn)
% 11.67/1.89  = { by axiom 1 (m__5309_1) }
% 11.67/1.89    szDzizrdt0(xd)
% 11.67/1.89  % SZS output end Proof
% 11.67/1.89  
% 11.67/1.89  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------