TSTP Solution File: RNG088+2 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : RNG088+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 : n031.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 13:59:13 EDT 2023

% Result   : Theorem 0.21s 0.56s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.12  % Problem  : RNG088+2 : TPTP v8.1.2. Released v4.0.0.
% 0.13/0.13  % 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 : n031.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 : Sun Aug 27 02:12:53 EDT 2023
% 0.14/0.34  % CPUTime  : 
% 0.21/0.56  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.21/0.56  
% 0.21/0.56  % SZS status Theorem
% 0.21/0.56  
% 0.21/0.56  % SZS output start Proof
% 0.21/0.56  Take the following subset of the input axioms:
% 0.21/0.56    fof(m__, conjecture, aElementOf0(sdtpldt0(xk, xm), xI) & aElementOf0(sdtpldt0(xl, xn), xJ)).
% 0.21/0.56    fof(m__870, hypothesis, aSet0(xI) & (![W0]: (aElementOf0(W0, xI) => (![W1]: (aElementOf0(W1, xI) => aElementOf0(sdtpldt0(W0, W1), xI)) & ![W1_2]: (aElement0(W1_2) => aElementOf0(sdtasdt0(W1_2, W0), xI)))) & (aIdeal0(xI) & (aSet0(xJ) & (![W0_2]: (aElementOf0(W0_2, xJ) => (![W1_2]: (aElementOf0(W1_2, xJ) => aElementOf0(sdtpldt0(W0_2, W1_2), xJ)) & ![W1_2]: (aElement0(W1_2) => aElementOf0(sdtasdt0(W1_2, W0_2), xJ)))) & aIdeal0(xJ)))))).
% 0.21/0.56    fof(m__934, hypothesis, aElementOf0(xk, xI) & (aElementOf0(xl, xJ) & xx=sdtpldt0(xk, xl))).
% 0.21/0.56    fof(m__967, hypothesis, aElementOf0(xm, xI) & (aElementOf0(xn, xJ) & xy=sdtpldt0(xm, xn))).
% 0.21/0.56  
% 0.21/0.56  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.56  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.56  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.56    fresh(y, y, x1...xn) = u
% 0.21/0.56    C => fresh(s, t, x1...xn) = v
% 0.21/0.56  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.56  variables of u and v.
% 0.21/0.56  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.56  input problem has no model of domain size 1).
% 0.21/0.56  
% 0.21/0.56  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.56  
% 0.21/0.56  Axiom 1 (m__934_1): aElementOf0(xk, xI) = true.
% 0.21/0.56  Axiom 2 (m__934_2): aElementOf0(xl, xJ) = true.
% 0.21/0.56  Axiom 3 (m__967_1): aElementOf0(xm, xI) = true.
% 0.21/0.56  Axiom 4 (m__967_2): aElementOf0(xn, xJ) = true.
% 0.21/0.56  Axiom 5 (m__870_6): fresh8(X, X, Y, Z) = true.
% 0.21/0.56  Axiom 6 (m__870_7): fresh6(X, X, Y, Z) = true.
% 0.21/0.56  Axiom 7 (m__870_6): fresh9(X, X, Y, Z) = aElementOf0(sdtpldt0(Y, Z), xI).
% 0.21/0.56  Axiom 8 (m__870_7): fresh7(X, X, Y, Z) = aElementOf0(sdtpldt0(Y, Z), xJ).
% 0.21/0.56  Axiom 9 (m__870_6): fresh9(aElementOf0(X, xI), true, Y, X) = fresh8(aElementOf0(Y, xI), true, Y, X).
% 0.21/0.56  Axiom 10 (m__870_7): fresh7(aElementOf0(X, xJ), true, Y, X) = fresh6(aElementOf0(Y, xJ), true, Y, X).
% 0.21/0.56  
% 0.21/0.56  Goal 1 (m__): tuple(aElementOf0(sdtpldt0(xk, xm), xI), aElementOf0(sdtpldt0(xl, xn), xJ)) = tuple(true, true).
% 0.21/0.56  Proof:
% 0.21/0.56    tuple(aElementOf0(sdtpldt0(xk, xm), xI), aElementOf0(sdtpldt0(xl, xn), xJ))
% 0.21/0.56  = { by axiom 7 (m__870_6) R->L }
% 0.21/0.56    tuple(fresh9(true, true, xk, xm), aElementOf0(sdtpldt0(xl, xn), xJ))
% 0.21/0.56  = { by axiom 8 (m__870_7) R->L }
% 0.21/0.56    tuple(fresh9(true, true, xk, xm), fresh7(true, true, xl, xn))
% 0.21/0.56  = { by axiom 3 (m__967_1) R->L }
% 0.21/0.56    tuple(fresh9(aElementOf0(xm, xI), true, xk, xm), fresh7(true, true, xl, xn))
% 0.21/0.56  = { by axiom 9 (m__870_6) }
% 0.21/0.56    tuple(fresh8(aElementOf0(xk, xI), true, xk, xm), fresh7(true, true, xl, xn))
% 0.21/0.56  = { by axiom 1 (m__934_1) }
% 0.21/0.56    tuple(fresh8(true, true, xk, xm), fresh7(true, true, xl, xn))
% 0.21/0.56  = { by axiom 5 (m__870_6) }
% 0.21/0.56    tuple(true, fresh7(true, true, xl, xn))
% 0.21/0.56  = { by axiom 4 (m__967_2) R->L }
% 0.21/0.56    tuple(true, fresh7(aElementOf0(xn, xJ), true, xl, xn))
% 0.21/0.56  = { by axiom 10 (m__870_7) }
% 0.21/0.56    tuple(true, fresh6(aElementOf0(xl, xJ), true, xl, xn))
% 0.21/0.56  = { by axiom 2 (m__934_2) }
% 0.21/0.56    tuple(true, fresh6(true, true, xl, xn))
% 0.21/0.56  = { by axiom 6 (m__870_7) }
% 0.21/0.56    tuple(true, true)
% 0.21/0.56  % SZS output end Proof
% 0.21/0.56  
% 0.21/0.56  RESULT: Theorem (the conjecture is true).
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