TSTP Solution File: NUM613+3 by Twee---2.5.0

View Problem - Process Solution

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% File     : Twee---2.5.0
% Problem  : NUM613+3 : TPTP v8.2.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : parallel-twee /export/starexec/sandbox2/benchmark/theBenchmark.p --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding

% Computer : n021.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 : Mon Jun 24 13:09:01 EDT 2024

% Result   : Theorem 52.04s 6.95s
% Output   : Proof 52.04s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.16  % Problem  : NUM613+3 : TPTP v8.2.0. Released v4.0.0.
% 0.10/0.16  % Command  : parallel-twee /export/starexec/sandbox2/benchmark/theBenchmark.p --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding
% 0.16/0.38  % Computer : n021.cluster.edu
% 0.16/0.38  % Model    : x86_64 x86_64
% 0.16/0.38  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.38  % Memory   : 8042.1875MB
% 0.16/0.38  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.16/0.38  % CPULimit : 300
% 0.16/0.38  % WCLimit  : 300
% 0.16/0.38  % DateTime : Sun Jun 23 02:58:24 EDT 2024
% 0.16/0.38  % CPUTime  : 
% 52.04/6.95  Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 52.04/6.95  
% 52.04/6.95  % SZS status Theorem
% 52.04/6.95  
% 52.04/6.96  % SZS output start Proof
% 52.04/6.96  Take the following subset of the input axioms:
% 52.04/6.96    fof(mSuccEquSucc, axiom, ![W0, W1]: ((aElementOf0(W0, szNzAzT0) & aElementOf0(W1, szNzAzT0)) => (szszuzczcdt0(W0)=szszuzczcdt0(W1) => W0=W1))).
% 52.04/6.96    fof(m__, conjecture, sbrdtbr0(xP)=xk).
% 52.04/6.96    fof(m__3533, hypothesis, aElementOf0(xk, szNzAzT0) & szszuzczcdt0(xk)=xK).
% 52.04/6.96    fof(m__5078, hypothesis, aSet0(xQ) & (![W0_2]: (aElementOf0(W0_2, xQ) => aElementOf0(W0_2, xO)) & (aSubsetOf0(xQ, xO) & (sbrdtbr0(xQ)=xK & aElementOf0(xQ, slbdtsldtrb0(xO, xK)))))).
% 52.04/6.96    fof(m__5255, hypothesis, sbrdtbr0(xQ)=szszuzczcdt0(xk) & (szszuzczcdt0(sbrdtbr0(xP))=sbrdtbr0(xQ) & aElementOf0(sbrdtbr0(xP), szNzAzT0))).
% 52.04/6.96  
% 52.04/6.96  Now clausify the problem and encode Horn clauses using encoding 3 of
% 52.04/6.96  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 52.04/6.96  We repeatedly replace C & s=t => u=v by the two clauses:
% 52.04/6.96    fresh(y, y, x1...xn) = u
% 52.04/6.96    C => fresh(s, t, x1...xn) = v
% 52.04/6.96  where fresh is a fresh function symbol and x1..xn are the free
% 52.04/6.96  variables of u and v.
% 52.04/6.96  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 52.04/6.96  input problem has no model of domain size 1).
% 52.04/6.96  
% 52.04/6.96  The encoding turns the above axioms into the following unit equations and goals:
% 52.04/6.96  
% 52.04/6.96  Axiom 1 (m__5078): sbrdtbr0(xQ) = xK.
% 52.04/6.96  Axiom 2 (m__3533): szszuzczcdt0(xk) = xK.
% 52.04/6.96  Axiom 3 (m__3533_1): aElementOf0(xk, szNzAzT0) = true2.
% 52.04/6.96  Axiom 4 (m__5255): szszuzczcdt0(sbrdtbr0(xP)) = sbrdtbr0(xQ).
% 52.04/6.96  Axiom 5 (m__5255_2): aElementOf0(sbrdtbr0(xP), szNzAzT0) = true2.
% 52.04/6.96  Axiom 6 (mSuccEquSucc): fresh473(X, X, Y, Z) = Z.
% 52.04/6.96  Axiom 7 (mSuccEquSucc): fresh14(X, X, Y, Z) = Y.
% 52.04/6.96  Axiom 8 (mSuccEquSucc): fresh472(X, X, Y, Z) = fresh473(szszuzczcdt0(Y), szszuzczcdt0(Z), Y, Z).
% 52.04/6.96  Axiom 9 (mSuccEquSucc): fresh472(aElementOf0(X, szNzAzT0), true2, Y, X) = fresh14(aElementOf0(Y, szNzAzT0), true2, Y, X).
% 52.04/6.96  
% 52.04/6.96  Goal 1 (m__): sbrdtbr0(xP) = xk.
% 52.04/6.96  Proof:
% 52.04/6.96    sbrdtbr0(xP)
% 52.04/6.96  = { by axiom 6 (mSuccEquSucc) R->L }
% 52.04/6.96    fresh473(xK, xK, xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 1 (m__5078) R->L }
% 52.04/6.96    fresh473(xK, sbrdtbr0(xQ), xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 4 (m__5255) R->L }
% 52.04/6.96    fresh473(xK, szszuzczcdt0(sbrdtbr0(xP)), xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 2 (m__3533) R->L }
% 52.04/6.96    fresh473(szszuzczcdt0(xk), szszuzczcdt0(sbrdtbr0(xP)), xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 8 (mSuccEquSucc) R->L }
% 52.04/6.96    fresh472(true2, true2, xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 5 (m__5255_2) R->L }
% 52.04/6.96    fresh472(aElementOf0(sbrdtbr0(xP), szNzAzT0), true2, xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 9 (mSuccEquSucc) }
% 52.04/6.96    fresh14(aElementOf0(xk, szNzAzT0), true2, xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 3 (m__3533_1) }
% 52.04/6.96    fresh14(true2, true2, xk, sbrdtbr0(xP))
% 52.04/6.96  = { by axiom 7 (mSuccEquSucc) }
% 52.04/6.96    xk
% 52.04/6.96  % SZS output end Proof
% 52.04/6.96  
% 52.04/6.96  RESULT: Theorem (the conjecture is true).
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