%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : NUM848+2 : TPTP v8.1.2. Released v4.1.0.
% 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 : n026.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 11:58:37 EDT 2023

% Result   : Theorem 0.14s 0.40s
% Output   : Proof 0.14s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : NUM848+2 : TPTP v8.1.2. Released v4.1.0.
% 0.00/0.14  % 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 : n026.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
% 0.14/0.36  % CPULimit : 300
% 0.14/0.36  % WCLimit  : 300
% 0.14/0.36  % DateTime : Fri Aug 25 13:10:02 EDT 2023
% 0.14/0.36  % CPUTime  : 
% 0.14/0.40  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.14/0.40  
% 0.14/0.40  % SZS status Theorem
% 0.14/0.40  
% 0.14/0.40  % SZS output start Proof
% 0.14/0.40  Take the following subset of the input axioms:
% 0.14/0.40    fof('holds(286, 441, 4)', conjecture, vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436))=vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439)))).
% 0.14/0.40    fof('qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))', axiom, ![Vd396, Vd397]: (vmul(Vd396, vsucc(Vd397))=vplus(vmul(Vd396, Vd397), Vd396) & vmul(Vd396, v1)=Vd396)).
% 0.14/0.40  
% 0.14/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.14/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.14/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.14/0.40    fresh(y, y, x1...xn) = u
% 0.14/0.40    C => fresh(s, t, x1...xn) = v
% 0.14/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.14/0.40  variables of u and v.
% 0.14/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.14/0.40  input problem has no model of domain size 1).
% 0.14/0.40  
% 0.14/0.40  The encoding turns the above axioms into the following unit equations and goals:
% 0.14/0.40  
% 0.14/0.41  Axiom 1 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))_1): vmul(X, vsucc(Y)) = vplus(vmul(X, Y), X).
% 0.14/0.41  
% 0.14/0.41  Goal 1 (holds(286, 441, 4)): vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436)) = vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439))).
% 0.14/0.41  Proof:
% 0.14/0.41    vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436))
% 0.14/0.41  = { by axiom 1 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))_1) R->L }
% 0.14/0.41    vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439)))
% 0.14/0.41  % SZS output end Proof
% 0.14/0.41  
% 0.14/0.41  RESULT: Theorem (the conjecture is true).
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