TSTP Solution File: NUM491+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM491+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 : n020.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 11:56:36 EDT 2023
% Result : Theorem 0.19s 0.52s
% Output : Proof 0.19s
% 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 : NUM491+3 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % 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 : n020.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 14:21:00 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.52 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.52
% 0.19/0.52 % SZS status Theorem
% 0.19/0.52
% 0.19/0.53 % SZS output start Proof
% 0.19/0.53 Take the following subset of the input axioms:
% 0.19/0.53 fof(mDefPrime, definition, ![W0]: (aNaturalNumber0(W0) => (isPrime0(W0) <=> (W0!=sz00 & (W0!=sz10 & ![W1]: ((aNaturalNumber0(W1) & doDivides0(W1, W0)) => (W1=sz10 | W1=W0))))))).
% 0.19/0.53 fof(m__, conjecture, ?[W0_2]: (aNaturalNumber0(W0_2) & sdtasdt0(xp, xm)=sdtasdt0(xp, W0_2)) | doDivides0(xp, sdtasdt0(xp, xm))).
% 0.19/0.53 fof(m__1799, hypothesis, ![W2, W0_2, W1_2]: ((aNaturalNumber0(W0_2) & (aNaturalNumber0(W1_2) & aNaturalNumber0(W2))) => ((((W2!=sz00 & (W2!=sz10 & ![W3]: ((aNaturalNumber0(W3) & (?[W4]: (aNaturalNumber0(W4) & W2=sdtasdt0(W3, W4)) & doDivides0(W3, W2))) => (W3=sz10 | W3=W2)))) | isPrime0(W2)) & (?[W3_2]: (aNaturalNumber0(W3_2) & sdtasdt0(W0_2, W1_2)=sdtasdt0(W2, W3_2)) | doDivides0(W2, sdtasdt0(W0_2, W1_2)))) => (iLess0(sdtpldt0(sdtpldt0(W0_2, W1_2), W2), sdtpldt0(sdtpldt0(xn, xm), xp)) => ((?[W3_2]: (aNaturalNumber0(W3_2) & W0_2=sdtasdt0(W2, W3_2)) & doDivides0(W2, W0_2)) | (?[W3_2]: (aNaturalNumber0(W3_2) & W1_2=sdtasdt0(W2, W3_2)) & doDivides0(W2, W1_2))))))).
% 0.19/0.53 fof(m__1837, hypothesis, aNaturalNumber0(xn) & (aNaturalNumber0(xm) & aNaturalNumber0(xp))).
% 0.19/0.53
% 0.19/0.53 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.53 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.53 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.53 fresh(y, y, x1...xn) = u
% 0.19/0.53 C => fresh(s, t, x1...xn) = v
% 0.19/0.53 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.53 variables of u and v.
% 0.19/0.53 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.53 input problem has no model of domain size 1).
% 0.19/0.53
% 0.19/0.53 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.53
% 0.19/0.53 Axiom 1 (m__1837_1): aNaturalNumber0(xm) = true2.
% 0.19/0.53
% 0.19/0.53 Goal 1 (m__): tuple3(sdtasdt0(xp, xm), aNaturalNumber0(X)) = tuple3(sdtasdt0(xp, X), true2).
% 0.19/0.53 The goal is true when:
% 0.19/0.53 X = xm
% 0.19/0.53
% 0.19/0.53 Proof:
% 0.19/0.53 tuple3(sdtasdt0(xp, xm), aNaturalNumber0(xm))
% 0.19/0.53 = { by axiom 1 (m__1837_1) }
% 0.19/0.53 tuple3(sdtasdt0(xp, xm), true2)
% 0.19/0.53 % SZS output end Proof
% 0.19/0.53
% 0.19/0.53 RESULT: Theorem (the conjecture is true).
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