TSTP Solution File: NUM480+2 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM480+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 : n019.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 11:56:31 EDT 2023
% Result : Theorem 8.39s 1.58s
% Output : Proof 8.39s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM480+2 : 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
% 0.13/0.34 % Computer : n019.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Fri Aug 25 17:23:58 EDT 2023
% 0.13/0.34 % CPUTime :
% 8.39/1.58 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 8.39/1.58
% 8.39/1.58 % SZS status Theorem
% 8.39/1.58
% 8.39/1.58 % SZS output start Proof
% 8.39/1.58 Take the following subset of the input axioms:
% 8.39/1.58 fof(mMulAsso, axiom, ![W0, W1, W2]: ((aNaturalNumber0(W0) & (aNaturalNumber0(W1) & aNaturalNumber0(W2))) => sdtasdt0(sdtasdt0(W0, W1), W2)=sdtasdt0(W0, sdtasdt0(W1, W2)))).
% 8.39/1.58 fof(m__, conjecture, (aNaturalNumber0(sdtsldt0(xm, xl)) & xm=sdtasdt0(xl, sdtsldt0(xm, xl))) => (sdtasdt0(xn, xm)=sdtasdt0(xl, sdtasdt0(xn, sdtsldt0(xm, xl))) | sdtasdt0(xn, sdtsldt0(xm, xl))=sdtsldt0(sdtasdt0(xn, xm), xl))).
% 8.39/1.58 fof(m__1524, hypothesis, aNaturalNumber0(xl) & aNaturalNumber0(xm)).
% 8.39/1.58 fof(m__1553, hypothesis, aNaturalNumber0(xn)).
% 8.39/1.58 fof(m__1594, hypothesis, aNaturalNumber0(sdtsldt0(xm, xl)) & (xm=sdtasdt0(xl, sdtsldt0(xm, xl)) & (aNaturalNumber0(sdtsldt0(sdtasdt0(xn, xm), xl)) & (sdtasdt0(xn, xm)=sdtasdt0(xl, sdtsldt0(sdtasdt0(xn, xm), xl)) & sdtasdt0(sdtasdt0(xl, xn), sdtsldt0(xm, xl))=sdtasdt0(xl, sdtsldt0(sdtasdt0(xn, xm), xl)))))).
% 8.39/1.58
% 8.39/1.58 Now clausify the problem and encode Horn clauses using encoding 3 of
% 8.39/1.58 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 8.39/1.58 We repeatedly replace C & s=t => u=v by the two clauses:
% 8.39/1.58 fresh(y, y, x1...xn) = u
% 8.39/1.58 C => fresh(s, t, x1...xn) = v
% 8.39/1.58 where fresh is a fresh function symbol and x1..xn are the free
% 8.39/1.58 variables of u and v.
% 8.39/1.58 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 8.39/1.58 input problem has no model of domain size 1).
% 8.39/1.58
% 8.39/1.58 The encoding turns the above axioms into the following unit equations and goals:
% 8.39/1.58
% 8.39/1.58 Axiom 1 (m__1524): aNaturalNumber0(xl) = true.
% 8.39/1.58 Axiom 2 (m__1553): aNaturalNumber0(xn) = true.
% 8.39/1.58 Axiom 3 (m__1594_4): aNaturalNumber0(sdtsldt0(xm, xl)) = true.
% 8.39/1.58 Axiom 4 (mMulAsso): fresh107(X, X, Y, Z, W) = sdtasdt0(Y, sdtasdt0(Z, W)).
% 8.39/1.58 Axiom 5 (mMulAsso): fresh21(X, X, Y, Z, W) = sdtasdt0(sdtasdt0(Y, Z), W).
% 8.39/1.58 Axiom 6 (m__1594_1): sdtasdt0(xn, xm) = sdtasdt0(xl, sdtsldt0(sdtasdt0(xn, xm), xl)).
% 8.39/1.58 Axiom 7 (m__1594): sdtasdt0(sdtasdt0(xl, xn), sdtsldt0(xm, xl)) = sdtasdt0(xl, sdtsldt0(sdtasdt0(xn, xm), xl)).
% 8.39/1.58 Axiom 8 (mMulAsso): fresh106(X, X, Y, Z, W) = fresh107(aNaturalNumber0(Y), true, Y, Z, W).
% 8.39/1.58 Axiom 9 (mMulAsso): fresh106(aNaturalNumber0(X), true, Y, Z, X) = fresh21(aNaturalNumber0(Z), true, Y, Z, X).
% 8.39/1.58
% 8.39/1.58 Goal 1 (m___2): sdtasdt0(xn, xm) = sdtasdt0(xl, sdtasdt0(xn, sdtsldt0(xm, xl))).
% 8.39/1.58 Proof:
% 8.39/1.58 sdtasdt0(xn, xm)
% 8.39/1.58 = { by axiom 6 (m__1594_1) }
% 8.39/1.58 sdtasdt0(xl, sdtsldt0(sdtasdt0(xn, xm), xl))
% 8.39/1.58 = { by axiom 7 (m__1594) R->L }
% 8.39/1.58 sdtasdt0(sdtasdt0(xl, xn), sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 5 (mMulAsso) R->L }
% 8.39/1.58 fresh21(true, true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 2 (m__1553) R->L }
% 8.39/1.58 fresh21(aNaturalNumber0(xn), true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 9 (mMulAsso) R->L }
% 8.39/1.58 fresh106(aNaturalNumber0(sdtsldt0(xm, xl)), true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 3 (m__1594_4) }
% 8.39/1.58 fresh106(true, true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 8 (mMulAsso) }
% 8.39/1.58 fresh107(aNaturalNumber0(xl), true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 1 (m__1524) }
% 8.39/1.58 fresh107(true, true, xl, xn, sdtsldt0(xm, xl))
% 8.39/1.58 = { by axiom 4 (mMulAsso) }
% 8.39/1.58 sdtasdt0(xl, sdtasdt0(xn, sdtsldt0(xm, xl)))
% 8.39/1.58 % SZS output end Proof
% 8.39/1.58
% 8.39/1.58 RESULT: Theorem (the conjecture is true).
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