TSTP Solution File: NUM525+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM525+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 : n014.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 11:56:46 EDT 2023
% Result : Theorem 26.38s 3.80s
% Output : Proof 26.38s
% 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 : NUM525+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
% 0.13/0.34 % Computer : n014.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 09:33:03 EDT 2023
% 0.13/0.35 % CPUTime :
% 26.38/3.80 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 26.38/3.80
% 26.38/3.80 % SZS status Theorem
% 26.38/3.80
% 26.38/3.80 % SZS output start Proof
% 26.38/3.80 Take the following subset of the input axioms:
% 26.38/3.80 fof(mMulAsso, axiom, ![W0, W1, W2]: ((aNaturalNumber0(W0) & (aNaturalNumber0(W1) & aNaturalNumber0(W2))) => sdtasdt0(sdtasdt0(W0, W1), W2)=sdtasdt0(W0, sdtasdt0(W1, W2)))).
% 26.38/3.80 fof(mMulComm, axiom, ![W0_2, W1_2]: ((aNaturalNumber0(W0_2) & aNaturalNumber0(W1_2)) => sdtasdt0(W0_2, W1_2)=sdtasdt0(W1_2, W0_2))).
% 26.38/3.80 fof(m__, conjecture, sdtasdt0(xp, sdtasdt0(xm, xm))=sdtasdt0(xp, sdtasdt0(xp, sdtasdt0(xq, xq)))).
% 26.38/3.80 fof(m__2987, hypothesis, aNaturalNumber0(xn) & (aNaturalNumber0(xm) & (aNaturalNumber0(xp) & (xn!=sz00 & (xm!=sz00 & xp!=sz00))))).
% 26.38/3.80 fof(m__3014, hypothesis, sdtasdt0(xp, sdtasdt0(xm, xm))=sdtasdt0(xn, xn)).
% 26.38/3.80 fof(m__3059, hypothesis, aNaturalNumber0(xq) & (xn=sdtasdt0(xp, xq) & xq=sdtsldt0(xn, xp))).
% 26.38/3.80
% 26.38/3.80 Now clausify the problem and encode Horn clauses using encoding 3 of
% 26.38/3.80 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 26.38/3.80 We repeatedly replace C & s=t => u=v by the two clauses:
% 26.38/3.80 fresh(y, y, x1...xn) = u
% 26.38/3.80 C => fresh(s, t, x1...xn) = v
% 26.38/3.80 where fresh is a fresh function symbol and x1..xn are the free
% 26.38/3.80 variables of u and v.
% 26.38/3.80 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 26.38/3.80 input problem has no model of domain size 1).
% 26.38/3.80
% 26.38/3.80 The encoding turns the above axioms into the following unit equations and goals:
% 26.38/3.80
% 26.38/3.80 Axiom 1 (m__2987_2): aNaturalNumber0(xp) = true2.
% 26.38/3.80 Axiom 2 (m__2987): aNaturalNumber0(xn) = true2.
% 26.38/3.80 Axiom 3 (m__3059_2): aNaturalNumber0(xq) = true2.
% 26.38/3.80 Axiom 4 (m__3059): xn = sdtasdt0(xp, xq).
% 26.38/3.80 Axiom 5 (m__3014): sdtasdt0(xp, sdtasdt0(xm, xm)) = sdtasdt0(xn, xn).
% 26.38/3.80 Axiom 6 (mMulComm): fresh20(X, X, Y, Z) = sdtasdt0(Y, Z).
% 26.38/3.80 Axiom 7 (mMulComm): fresh19(X, X, Y, Z) = sdtasdt0(Z, Y).
% 26.38/3.80 Axiom 8 (mMulAsso): fresh107(X, X, Y, Z, W) = sdtasdt0(Y, sdtasdt0(Z, W)).
% 26.38/3.80 Axiom 9 (mMulAsso): fresh21(X, X, Y, Z, W) = sdtasdt0(sdtasdt0(Y, Z), W).
% 26.38/3.80 Axiom 10 (mMulComm): fresh20(aNaturalNumber0(X), true2, Y, X) = fresh19(aNaturalNumber0(Y), true2, Y, X).
% 26.38/3.80 Axiom 11 (mMulAsso): fresh106(X, X, Y, Z, W) = fresh107(aNaturalNumber0(Y), true2, Y, Z, W).
% 26.38/3.80 Axiom 12 (mMulAsso): fresh106(aNaturalNumber0(X), true2, Y, Z, X) = fresh21(aNaturalNumber0(Z), true2, Y, Z, X).
% 26.38/3.80
% 26.38/3.80 Lemma 13: fresh106(X, X, xp, Y, Z) = sdtasdt0(xp, sdtasdt0(Y, Z)).
% 26.38/3.80 Proof:
% 26.38/3.80 fresh106(X, X, xp, Y, Z)
% 26.38/3.80 = { by axiom 11 (mMulAsso) }
% 26.38/3.80 fresh107(aNaturalNumber0(xp), true2, xp, Y, Z)
% 26.38/3.80 = { by axiom 1 (m__2987_2) }
% 26.38/3.80 fresh107(true2, true2, xp, Y, Z)
% 26.38/3.80 = { by axiom 8 (mMulAsso) }
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(Y, Z))
% 26.38/3.80
% 26.38/3.80 Lemma 14: fresh106(aNaturalNumber0(X), true2, Y, xq, X) = sdtasdt0(sdtasdt0(Y, xq), X).
% 26.38/3.80 Proof:
% 26.38/3.80 fresh106(aNaturalNumber0(X), true2, Y, xq, X)
% 26.38/3.80 = { by axiom 12 (mMulAsso) }
% 26.38/3.80 fresh21(aNaturalNumber0(xq), true2, Y, xq, X)
% 26.38/3.80 = { by axiom 3 (m__3059_2) }
% 26.38/3.80 fresh21(true2, true2, Y, xq, X)
% 26.38/3.80 = { by axiom 9 (mMulAsso) }
% 26.38/3.80 sdtasdt0(sdtasdt0(Y, xq), X)
% 26.38/3.80
% 26.38/3.80 Goal 1 (m__): sdtasdt0(xp, sdtasdt0(xm, xm)) = sdtasdt0(xp, sdtasdt0(xp, sdtasdt0(xq, xq))).
% 26.38/3.80 Proof:
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(xm, xm))
% 26.38/3.80 = { by axiom 5 (m__3014) }
% 26.38/3.80 sdtasdt0(xn, xn)
% 26.38/3.80 = { by axiom 4 (m__3059) }
% 26.38/3.80 sdtasdt0(sdtasdt0(xp, xq), xn)
% 26.38/3.80 = { by lemma 14 R->L }
% 26.38/3.80 fresh106(aNaturalNumber0(xn), true2, xp, xq, xn)
% 26.38/3.80 = { by axiom 2 (m__2987) }
% 26.38/3.80 fresh106(true2, true2, xp, xq, xn)
% 26.38/3.80 = { by lemma 13 }
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(xq, xn))
% 26.38/3.80 = { by axiom 7 (mMulComm) R->L }
% 26.38/3.80 sdtasdt0(xp, fresh19(true2, true2, xn, xq))
% 26.38/3.80 = { by axiom 2 (m__2987) R->L }
% 26.38/3.80 sdtasdt0(xp, fresh19(aNaturalNumber0(xn), true2, xn, xq))
% 26.38/3.80 = { by axiom 10 (mMulComm) R->L }
% 26.38/3.80 sdtasdt0(xp, fresh20(aNaturalNumber0(xq), true2, xn, xq))
% 26.38/3.80 = { by axiom 3 (m__3059_2) }
% 26.38/3.80 sdtasdt0(xp, fresh20(true2, true2, xn, xq))
% 26.38/3.80 = { by axiom 6 (mMulComm) }
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(xn, xq))
% 26.38/3.80 = { by axiom 4 (m__3059) }
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(sdtasdt0(xp, xq), xq))
% 26.38/3.80 = { by lemma 14 R->L }
% 26.38/3.80 sdtasdt0(xp, fresh106(aNaturalNumber0(xq), true2, xp, xq, xq))
% 26.38/3.80 = { by axiom 3 (m__3059_2) }
% 26.38/3.80 sdtasdt0(xp, fresh106(true2, true2, xp, xq, xq))
% 26.38/3.80 = { by lemma 13 }
% 26.38/3.80 sdtasdt0(xp, sdtasdt0(xp, sdtasdt0(xq, xq)))
% 26.38/3.80 % SZS output end Proof
% 26.38/3.80
% 26.38/3.80 RESULT: Theorem (the conjecture is true).
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