TSTP Solution File: NUM514+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM514+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 11:56:43 EDT 2023
% Result : Theorem 0.20s 0.54s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : NUM514+3 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35 % Computer : n020.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Fri Aug 25 16:04:15 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.20/0.54 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.54
% 0.20/0.54 % SZS status Theorem
% 0.20/0.54
% 0.20/0.54 % SZS output start Proof
% 0.20/0.54 Take the following subset of the input axioms:
% 0.20/0.55 fof(mDefPrime, definition, ![W0]: (aNaturalNumber0(W0) => (isPrime0(W0) <=> (W0!=sz00 & (W0!=sz10 & ![W1]: ((aNaturalNumber0(W1) & doDivides0(W1, W0)) => (W1=sz10 | W1=W0))))))).
% 0.20/0.55 fof(m__, conjecture, (aNaturalNumber0(sdtsldt0(xn, xr)) & xn=sdtasdt0(xr, sdtsldt0(xn, xr))) => (?[W0_2]: (aNaturalNumber0(W0_2) & sdtasdt0(sdtsldt0(xn, xr), xm)=sdtasdt0(xp, W0_2)) | doDivides0(xp, sdtasdt0(sdtsldt0(xn, xr), xm)))).
% 0.20/0.55 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))))))).
% 0.20/0.55 fof(m__1870, hypothesis, ~(?[W0_2]: (aNaturalNumber0(W0_2) & sdtpldt0(xp, W0_2)=xn) | sdtlseqdt0(xp, xn))).
% 0.20/0.55 fof(m__2075, hypothesis, ~(?[W0_2]: (aNaturalNumber0(W0_2) & sdtpldt0(xp, W0_2)=xm) | sdtlseqdt0(xp, xm))).
% 0.20/0.55 fof(m__2613, hypothesis, aNaturalNumber0(sdtsldt0(xk, xr)) & (xk=sdtasdt0(xr, sdtsldt0(xk, xr)) & (aNaturalNumber0(sdtsldt0(xn, xr)) & (xn=sdtasdt0(xr, sdtsldt0(xn, xr)) & sdtasdt0(xp, sdtsldt0(xk, xr))=sdtasdt0(sdtsldt0(xn, xr), xm))))).
% 0.20/0.55
% 0.20/0.55 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.55 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.55 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.55 fresh(y, y, x1...xn) = u
% 0.20/0.55 C => fresh(s, t, x1...xn) = v
% 0.20/0.55 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.55 variables of u and v.
% 0.20/0.55 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.55 input problem has no model of domain size 1).
% 0.20/0.55
% 0.20/0.55 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.55
% 0.20/0.55 Axiom 1 (m__2613_4): aNaturalNumber0(sdtsldt0(xk, xr)) = true2.
% 0.20/0.55 Axiom 2 (m__2613): sdtasdt0(xp, sdtsldt0(xk, xr)) = sdtasdt0(sdtsldt0(xn, xr), xm).
% 0.20/0.55
% 0.20/0.55 Goal 1 (m___2): tuple3(sdtasdt0(sdtsldt0(xn, xr), xm), aNaturalNumber0(X)) = tuple3(sdtasdt0(xp, X), true2).
% 0.20/0.55 The goal is true when:
% 0.20/0.55 X = sdtsldt0(xk, xr)
% 0.20/0.55
% 0.20/0.55 Proof:
% 0.20/0.55 tuple3(sdtasdt0(sdtsldt0(xn, xr), xm), aNaturalNumber0(sdtsldt0(xk, xr)))
% 0.20/0.55 = { by axiom 2 (m__2613) R->L }
% 0.20/0.55 tuple3(sdtasdt0(xp, sdtsldt0(xk, xr)), aNaturalNumber0(sdtsldt0(xk, xr)))
% 0.20/0.55 = { by axiom 1 (m__2613_4) }
% 0.20/0.55 tuple3(sdtasdt0(xp, sdtsldt0(xk, xr)), true2)
% 0.20/0.55 % SZS output end Proof
% 0.20/0.55
% 0.20/0.55 RESULT: Theorem (the conjecture is true).
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