TSTP Solution File: NUM603+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM603+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 : n001.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 11:57:13 EDT 2023
% Result : Theorem 5.67s 1.13s
% Output : Proof 5.67s
% 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 : NUM603+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
% 0.13/0.34 % Computer : n001.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 10:25:58 EDT 2023
% 0.13/0.34 % CPUTime :
% 5.67/1.13 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 5.67/1.13
% 5.67/1.13 % SZS status Theorem
% 5.67/1.13
% 5.67/1.14 % SZS output start Proof
% 5.67/1.14 Take the following subset of the input axioms:
% 5.67/1.14 fof(mZeroLess, axiom, ![W0]: (aElementOf0(W0, szNzAzT0) => sdtlseqdt0(sz00, W0))).
% 5.67/1.14 fof(mZeroNum, axiom, aElementOf0(sz00, szNzAzT0)).
% 5.67/1.14 fof(m__, conjecture, aSubsetOf0(sdtlpdtrp0(xN, xi), xS)).
% 5.67/1.14 fof(m__3623, hypothesis, aFunction0(xN) & (szDzozmdt0(xN)=szNzAzT0 & (sdtlpdtrp0(xN, sz00)=xS & ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ((aSubsetOf0(sdtlpdtrp0(xN, W0_2), szNzAzT0) & isCountable0(sdtlpdtrp0(xN, W0_2))) => (aSubsetOf0(sdtlpdtrp0(xN, szszuzczcdt0(W0_2)), sdtmndt0(sdtlpdtrp0(xN, W0_2), szmzizndt0(sdtlpdtrp0(xN, W0_2)))) & isCountable0(sdtlpdtrp0(xN, szszuzczcdt0(W0_2))))))))).
% 5.67/1.14 fof(m__3754, hypothesis, ![W1, W0_2]: ((aElementOf0(W0_2, szNzAzT0) & aElementOf0(W1, szNzAzT0)) => (sdtlseqdt0(W1, W0_2) => aSubsetOf0(sdtlpdtrp0(xN, W0_2), sdtlpdtrp0(xN, W1))))).
% 5.67/1.14 fof(m__5034, hypothesis, aElementOf0(xi, szNzAzT0) & sdtlpdtrp0(xe, xi)=xx).
% 5.67/1.14
% 5.67/1.14 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.67/1.14 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.67/1.14 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.67/1.14 fresh(y, y, x1...xn) = u
% 5.67/1.14 C => fresh(s, t, x1...xn) = v
% 5.67/1.14 where fresh is a fresh function symbol and x1..xn are the free
% 5.67/1.14 variables of u and v.
% 5.67/1.14 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.67/1.14 input problem has no model of domain size 1).
% 5.67/1.14
% 5.67/1.14 The encoding turns the above axioms into the following unit equations and goals:
% 5.67/1.14
% 5.67/1.14 Axiom 1 (mZeroNum): aElementOf0(sz00, szNzAzT0) = true2.
% 5.67/1.14 Axiom 2 (m__5034_1): aElementOf0(xi, szNzAzT0) = true2.
% 5.67/1.14 Axiom 3 (m__3623_1): sdtlpdtrp0(xN, sz00) = xS.
% 5.67/1.14 Axiom 4 (mZeroLess): fresh27(X, X, Y) = true2.
% 5.67/1.14 Axiom 5 (m__3754): fresh135(X, X, Y, Z) = true2.
% 5.67/1.14 Axiom 6 (mZeroLess): fresh27(aElementOf0(X, szNzAzT0), true2, X) = sdtlseqdt0(sz00, X).
% 5.67/1.14 Axiom 7 (m__3754): fresh23(X, X, Y, Z) = aSubsetOf0(sdtlpdtrp0(xN, Y), sdtlpdtrp0(xN, Z)).
% 5.67/1.14 Axiom 8 (m__3754): fresh134(X, X, Y, Z) = fresh135(aElementOf0(Y, szNzAzT0), true2, Y, Z).
% 5.67/1.14 Axiom 9 (m__3754): fresh134(sdtlseqdt0(X, Y), true2, Y, X) = fresh23(aElementOf0(X, szNzAzT0), true2, Y, X).
% 5.67/1.14
% 5.67/1.14 Goal 1 (m__): aSubsetOf0(sdtlpdtrp0(xN, xi), xS) = true2.
% 5.67/1.14 Proof:
% 5.67/1.14 aSubsetOf0(sdtlpdtrp0(xN, xi), xS)
% 5.67/1.14 = { by axiom 3 (m__3623_1) R->L }
% 5.67/1.14 aSubsetOf0(sdtlpdtrp0(xN, xi), sdtlpdtrp0(xN, sz00))
% 5.67/1.14 = { by axiom 7 (m__3754) R->L }
% 5.67/1.14 fresh23(true2, true2, xi, sz00)
% 5.67/1.14 = { by axiom 1 (mZeroNum) R->L }
% 5.67/1.14 fresh23(aElementOf0(sz00, szNzAzT0), true2, xi, sz00)
% 5.67/1.14 = { by axiom 9 (m__3754) R->L }
% 5.67/1.14 fresh134(sdtlseqdt0(sz00, xi), true2, xi, sz00)
% 5.67/1.14 = { by axiom 6 (mZeroLess) R->L }
% 5.67/1.14 fresh134(fresh27(aElementOf0(xi, szNzAzT0), true2, xi), true2, xi, sz00)
% 5.67/1.14 = { by axiom 2 (m__5034_1) }
% 5.67/1.14 fresh134(fresh27(true2, true2, xi), true2, xi, sz00)
% 5.67/1.14 = { by axiom 4 (mZeroLess) }
% 5.67/1.14 fresh134(true2, true2, xi, sz00)
% 5.67/1.14 = { by axiom 8 (m__3754) }
% 5.67/1.14 fresh135(aElementOf0(xi, szNzAzT0), true2, xi, sz00)
% 5.67/1.14 = { by axiom 2 (m__5034_1) }
% 5.67/1.14 fresh135(true2, true2, xi, sz00)
% 5.67/1.14 = { by axiom 5 (m__3754) }
% 5.67/1.14 true2
% 5.67/1.14 % SZS output end Proof
% 5.67/1.14
% 5.67/1.14 RESULT: Theorem (the conjecture is true).
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