TSTP Solution File: NUM561+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM561+2 : 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 : n011.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 11:56:59 EDT 2023
% Result : Theorem 4.20s 0.87s
% Output : Proof 4.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.11 % Problem : NUM561+2 : TPTP v8.1.2. Released v4.0.0.
% 0.09/0.11 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.10/0.32 % Computer : n011.cluster.edu
% 0.10/0.32 % Model : x86_64 x86_64
% 0.10/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.10/0.32 % Memory : 8042.1875MB
% 0.10/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.10/0.32 % CPULimit : 300
% 0.10/0.32 % WCLimit : 300
% 0.10/0.32 % DateTime : Fri Aug 25 14:05:53 EDT 2023
% 0.10/0.32 % CPUTime :
% 4.20/0.87 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 4.20/0.87
% 4.20/0.87 % SZS status Theorem
% 4.20/0.87
% 4.20/0.87 % SZS output start Proof
% 4.20/0.87 Take the following subset of the input axioms:
% 4.20/0.87 fof(mCountNFin, axiom, ![W0]: ((aSet0(W0) & isCountable0(W0)) => ~isFinite0(W0))).
% 4.20/0.87 fof(mCountNFin_01, axiom, ![W0_2]: ((aSet0(W0_2) & isCountable0(W0_2)) => W0_2!=slcrc0)).
% 4.20/0.87 fof(mDefDiff, definition, ![W1, W0_2]: ((aSet0(W0_2) & aElement0(W1)) => ![W2]: (W2=sdtmndt0(W0_2, W1) <=> (aSet0(W2) & ![W3]: (aElementOf0(W3, W2) <=> (aElement0(W3) & (aElementOf0(W3, W0_2) & W3!=W1))))))).
% 4.20/0.87 fof(mDefEmp, definition, ![W0_2]: (W0_2=slcrc0 <=> (aSet0(W0_2) & ~?[W1_2]: aElementOf0(W1_2, W0_2)))).
% 4.20/0.87 fof(mNatNSucc, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => W0_2!=szszuzczcdt0(W0_2))).
% 4.20/0.87 fof(mNoScLessZr, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ~sdtlseqdt0(szszuzczcdt0(W0_2), sz00))).
% 4.20/0.87 fof(mSuccNum, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => (aElementOf0(szszuzczcdt0(W0_2), szNzAzT0) & szszuzczcdt0(W0_2)!=sz00))).
% 4.20/0.88 fof(m__, conjecture, ?[W0_2]: (aElementOf0(W0_2, szDzozmdt0(xF)) & sdtlpdtrp0(xF, W0_2)=sdtlpdtrp0(xF, xx)) | aElementOf0(sdtlpdtrp0(xF, xx), sdtlcdtrc0(xF, szDzozmdt0(xF)))).
% 4.20/0.88 fof(m__2911_02, hypothesis, aElementOf0(xx, szDzozmdt0(xF))).
% 4.20/0.88
% 4.20/0.88 Now clausify the problem and encode Horn clauses using encoding 3 of
% 4.20/0.88 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 4.20/0.88 We repeatedly replace C & s=t => u=v by the two clauses:
% 4.20/0.88 fresh(y, y, x1...xn) = u
% 4.20/0.88 C => fresh(s, t, x1...xn) = v
% 4.20/0.88 where fresh is a fresh function symbol and x1..xn are the free
% 4.20/0.88 variables of u and v.
% 4.20/0.88 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 4.20/0.88 input problem has no model of domain size 1).
% 4.20/0.88
% 4.20/0.88 The encoding turns the above axioms into the following unit equations and goals:
% 4.20/0.88
% 4.20/0.88 Axiom 1 (m__2911_02): aElementOf0(xx, szDzozmdt0(xF)) = true2.
% 4.20/0.88
% 4.20/0.88 Goal 1 (m__): tuple4(sdtlpdtrp0(xF, X), aElementOf0(X, szDzozmdt0(xF))) = tuple4(sdtlpdtrp0(xF, xx), true2).
% 4.20/0.88 The goal is true when:
% 4.20/0.88 X = xx
% 4.20/0.88
% 4.20/0.88 Proof:
% 4.20/0.88 tuple4(sdtlpdtrp0(xF, xx), aElementOf0(xx, szDzozmdt0(xF)))
% 4.20/0.88 = { by axiom 1 (m__2911_02) }
% 4.20/0.88 tuple4(sdtlpdtrp0(xF, xx), true2)
% 4.20/0.88 % SZS output end Proof
% 4.20/0.88
% 4.20/0.88 RESULT: Theorem (the conjecture is true).
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