TSTP Solution File: NUM429+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM429+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 : n024.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 11:56:14 EDT 2023
% Result : Theorem 0.20s 0.45s
% 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
% 0.11/0.13 % Problem : NUM429+3 : TPTP v8.1.2. Released v4.0.0.
% 0.11/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n024.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Fri Aug 25 08:12:39 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.45 Command-line arguments: --no-flatten-goal
% 0.20/0.45
% 0.20/0.45 % SZS status Theorem
% 0.20/0.45
% 0.20/0.45 % SZS output start Proof
% 0.20/0.45 Take the following subset of the input axioms:
% 0.20/0.45 fof(mDivisor, definition, ![W0]: (aInteger0(W0) => ![W1]: (aDivisorOf0(W1, W0) <=> (aInteger0(W1) & (W1!=sz00 & ?[W2]: (aInteger0(W2) & sdtasdt0(W1, W2)=W0)))))).
% 0.20/0.45 fof(m__, conjecture, ?[W0_2]: (aInteger0(W0_2) & sdtasdt0(xq, W0_2)=sdtpldt0(xa, smndt0(xb)))).
% 0.20/0.45 fof(m__853, hypothesis, ?[W0_2]: (aInteger0(W0_2) & sdtasdt0(xq, W0_2)=sdtpldt0(xa, smndt0(xb))) & (aDivisorOf0(xq, sdtpldt0(xa, smndt0(xb))) & (sdteqdtlpzmzozddtrp0(xa, xb, xq) & (?[W0_2]: (aInteger0(W0_2) & sdtasdt0(xq, W0_2)=sdtpldt0(xb, smndt0(xc))) & (aDivisorOf0(xq, sdtpldt0(xb, smndt0(xc))) & sdteqdtlpzmzozddtrp0(xb, xc, xq)))))).
% 0.20/0.45
% 0.20/0.45 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.45 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.45 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.45 fresh(y, y, x1...xn) = u
% 0.20/0.45 C => fresh(s, t, x1...xn) = v
% 0.20/0.45 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.45 variables of u and v.
% 0.20/0.45 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.45 input problem has no model of domain size 1).
% 0.20/0.45
% 0.20/0.45 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.45
% 0.20/0.45 Axiom 1 (m__853_2): aInteger0(w0_2) = true2.
% 0.20/0.45 Axiom 2 (m__853): sdtasdt0(xq, w0_2) = sdtpldt0(xa, smndt0(xb)).
% 0.20/0.45
% 0.20/0.45 Goal 1 (m__): tuple2(sdtasdt0(xq, X), aInteger0(X)) = tuple2(sdtpldt0(xa, smndt0(xb)), true2).
% 0.20/0.45 The goal is true when:
% 0.20/0.45 X = w0_2
% 0.20/0.45
% 0.20/0.45 Proof:
% 0.20/0.45 tuple2(sdtasdt0(xq, w0_2), aInteger0(w0_2))
% 0.20/0.45 = { by axiom 1 (m__853_2) }
% 0.20/0.45 tuple2(sdtasdt0(xq, w0_2), true2)
% 0.20/0.45 = { by axiom 2 (m__853) }
% 0.20/0.45 tuple2(sdtpldt0(xa, smndt0(xb)), true2)
% 0.20/0.45 % SZS output end Proof
% 0.20/0.45
% 0.20/0.45 RESULT: Theorem (the conjecture is true).
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