TSTP Solution File: NUM472+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM472+1 : 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 : n027.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 11:56:28 EDT 2023
% Result : Theorem 40.57s 5.88s
% Output : Proof 40.57s
% 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 : NUM472+1 : 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
% 0.11/0.33 % Computer : n027.cluster.edu
% 0.11/0.33 % Model : x86_64 x86_64
% 0.11/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.33 % Memory : 8042.1875MB
% 0.11/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.33 % CPULimit : 300
% 0.11/0.33 % WCLimit : 300
% 0.11/0.33 % DateTime : Fri Aug 25 16:02:33 EDT 2023
% 0.11/0.34 % CPUTime :
% 40.57/5.88 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 40.57/5.88
% 40.57/5.88 % SZS status Theorem
% 40.57/5.88
% 40.57/5.89 % SZS output start Proof
% 40.57/5.89 Take the following subset of the input axioms:
% 40.57/5.89 fof(mDefLE, definition, ![W0, W1]: ((aNaturalNumber0(W0) & aNaturalNumber0(W1)) => (sdtlseqdt0(W0, W1) <=> ?[W2]: (aNaturalNumber0(W2) & sdtpldt0(W0, W2)=W1)))).
% 40.57/5.89 fof(mSortsB, axiom, ![W0_2, W1_2]: ((aNaturalNumber0(W0_2) & aNaturalNumber0(W1_2)) => aNaturalNumber0(sdtpldt0(W0_2, W1_2)))).
% 40.57/5.89 fof(m__, conjecture, sdtlseqdt0(xm, sdtpldt0(xm, xn))).
% 40.57/5.89 fof(m__1324, hypothesis, aNaturalNumber0(xl) & (aNaturalNumber0(xm) & aNaturalNumber0(xn))).
% 40.57/5.89
% 40.57/5.89 Now clausify the problem and encode Horn clauses using encoding 3 of
% 40.57/5.89 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 40.57/5.89 We repeatedly replace C & s=t => u=v by the two clauses:
% 40.57/5.89 fresh(y, y, x1...xn) = u
% 40.57/5.89 C => fresh(s, t, x1...xn) = v
% 40.57/5.89 where fresh is a fresh function symbol and x1..xn are the free
% 40.57/5.89 variables of u and v.
% 40.57/5.89 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 40.57/5.89 input problem has no model of domain size 1).
% 40.57/5.89
% 40.57/5.89 The encoding turns the above axioms into the following unit equations and goals:
% 40.57/5.89
% 40.57/5.89 Axiom 1 (m__1324_1): aNaturalNumber0(xm) = true.
% 40.57/5.89 Axiom 2 (m__1324_2): aNaturalNumber0(xn) = true.
% 40.57/5.89 Axiom 3 (mDefLE): fresh82(X, X, Y, Z) = true.
% 40.57/5.89 Axiom 4 (mSortsB): fresh18(X, X, Y, Z) = aNaturalNumber0(sdtpldt0(Y, Z)).
% 40.57/5.89 Axiom 5 (mSortsB): fresh17(X, X, Y, Z) = true.
% 40.57/5.89 Axiom 6 (mDefLE): fresh80(X, X, Y, Z, W) = sdtlseqdt0(Y, Z).
% 40.57/5.89 Axiom 7 (mSortsB): fresh18(aNaturalNumber0(X), true, Y, X) = fresh17(aNaturalNumber0(Y), true, Y, X).
% 40.57/5.89 Axiom 8 (mDefLE): fresh81(X, X, Y, Z, W) = fresh82(sdtpldt0(Y, W), Z, Y, Z).
% 40.57/5.89 Axiom 9 (mDefLE): fresh79(X, X, Y, Z, W) = fresh80(aNaturalNumber0(Y), true, Y, Z, W).
% 40.57/5.89 Axiom 10 (mDefLE): fresh79(aNaturalNumber0(X), true, Y, Z, X) = fresh81(aNaturalNumber0(Z), true, Y, Z, X).
% 40.57/5.89
% 40.57/5.89 Goal 1 (m__): sdtlseqdt0(xm, sdtpldt0(xm, xn)) = true.
% 40.57/5.89 Proof:
% 40.57/5.89 sdtlseqdt0(xm, sdtpldt0(xm, xn))
% 40.57/5.89 = { by axiom 6 (mDefLE) R->L }
% 40.57/5.89 fresh80(true, true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 1 (m__1324_1) R->L }
% 40.57/5.89 fresh80(aNaturalNumber0(xm), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 9 (mDefLE) R->L }
% 40.57/5.89 fresh79(true, true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 2 (m__1324_2) R->L }
% 40.57/5.89 fresh79(aNaturalNumber0(xn), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 10 (mDefLE) }
% 40.57/5.89 fresh81(aNaturalNumber0(sdtpldt0(xm, xn)), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 4 (mSortsB) R->L }
% 40.57/5.89 fresh81(fresh18(true, true, xm, xn), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 2 (m__1324_2) R->L }
% 40.57/5.89 fresh81(fresh18(aNaturalNumber0(xn), true, xm, xn), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 7 (mSortsB) }
% 40.57/5.89 fresh81(fresh17(aNaturalNumber0(xm), true, xm, xn), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 1 (m__1324_1) }
% 40.57/5.89 fresh81(fresh17(true, true, xm, xn), true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 5 (mSortsB) }
% 40.57/5.89 fresh81(true, true, xm, sdtpldt0(xm, xn), xn)
% 40.57/5.89 = { by axiom 8 (mDefLE) }
% 40.57/5.89 fresh82(sdtpldt0(xm, xn), sdtpldt0(xm, xn), xm, sdtpldt0(xm, xn))
% 40.57/5.89 = { by axiom 3 (mDefLE) }
% 40.57/5.89 true
% 40.57/5.89 % SZS output end Proof
% 40.57/5.89
% 40.57/5.89 RESULT: Theorem (the conjecture is true).
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