TSTP Solution File: NUM582+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM582+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 : n023.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 11:57:06 EDT 2023
% Result : Theorem 8.07s 1.42s
% Output : Proof 8.07s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : NUM582+3 : TPTP v8.1.2. Released v4.0.0.
% 0.04/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33 % Computer : n023.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Fri Aug 25 16:09:11 EDT 2023
% 0.12/0.34 % CPUTime :
% 8.07/1.42 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 8.07/1.42
% 8.07/1.42 % SZS status Theorem
% 8.07/1.42
% 8.07/1.42 % SZS output start Proof
% 8.07/1.42 Take the following subset of the input axioms:
% 8.07/1.43 fof(mZeroLess, axiom, ![W0]: (aElementOf0(W0, szNzAzT0) => sdtlseqdt0(sz00, W0))).
% 8.07/1.43 fof(mZeroNum, axiom, aElementOf0(sz00, szNzAzT0)).
% 8.07/1.43 fof(m__, conjecture, ![W0_2]: (aElementOf0(W0_2, sdtlpdtrp0(xN, xi)) => aElementOf0(W0_2, xS)) | aSubsetOf0(sdtlpdtrp0(xN, xi), xS)).
% 8.07/1.43 fof(m__3623, hypothesis, aFunction0(xN) & (szDzozmdt0(xN)=szNzAzT0 & (sdtlpdtrp0(xN, sz00)=xS & ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ((((aSet0(sdtlpdtrp0(xN, W0_2)) & ![W1]: (aElementOf0(W1, sdtlpdtrp0(xN, W0_2)) => aElementOf0(W1, szNzAzT0))) | aSubsetOf0(sdtlpdtrp0(xN, W0_2), szNzAzT0)) & isCountable0(sdtlpdtrp0(xN, W0_2))) => (aElementOf0(szmzizndt0(sdtlpdtrp0(xN, W0_2)), sdtlpdtrp0(xN, W0_2)) & (![W1_2]: (aElementOf0(W1_2, sdtlpdtrp0(xN, W0_2)) => sdtlseqdt0(szmzizndt0(sdtlpdtrp0(xN, W0_2)), W1_2)) & (aSet0(sdtmndt0(sdtlpdtrp0(xN, W0_2), szmzizndt0(sdtlpdtrp0(xN, W0_2)))) & (![W1_2]: (aElementOf0(W1_2, sdtmndt0(sdtlpdtrp0(xN, W0_2), szmzizndt0(sdtlpdtrp0(xN, W0_2)))) <=> (aElement0(W1_2) & (aElementOf0(W1_2, sdtlpdtrp0(xN, W0_2)) & W1_2!=szmzizndt0(sdtlpdtrp0(xN, W0_2))))) & (aSet0(sdtlpdtrp0(xN, szszuzczcdt0(W0_2))) & (![W1_2]: (aElementOf0(W1_2, sdtlpdtrp0(xN, szszuzczcdt0(W0_2))) => aElementOf0(W1_2, sdtmndt0(sdtlpdtrp0(xN, W0_2), szmzizndt0(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))))))))))))))).
% 8.07/1.43 fof(m__3754, hypothesis, ![W0_2, W1_2]: ((aElementOf0(W0_2, szNzAzT0) & aElementOf0(W1_2, szNzAzT0)) => (sdtlseqdt0(W1_2, W0_2) => (![W2]: (aElementOf0(W2, sdtlpdtrp0(xN, W0_2)) => aElementOf0(W2, sdtlpdtrp0(xN, W1_2))) & aSubsetOf0(sdtlpdtrp0(xN, W0_2), sdtlpdtrp0(xN, W1_2)))))).
% 8.07/1.43 fof(m__3989, hypothesis, aElementOf0(xi, szNzAzT0)).
% 8.07/1.43
% 8.07/1.43 Now clausify the problem and encode Horn clauses using encoding 3 of
% 8.07/1.43 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 8.07/1.43 We repeatedly replace C & s=t => u=v by the two clauses:
% 8.07/1.43 fresh(y, y, x1...xn) = u
% 8.07/1.43 C => fresh(s, t, x1...xn) = v
% 8.07/1.43 where fresh is a fresh function symbol and x1..xn are the free
% 8.07/1.43 variables of u and v.
% 8.07/1.43 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 8.07/1.43 input problem has no model of domain size 1).
% 8.07/1.43
% 8.07/1.43 The encoding turns the above axioms into the following unit equations and goals:
% 8.07/1.43
% 8.07/1.43 Axiom 1 (m__3989): aElementOf0(xi, szNzAzT0) = true2.
% 8.07/1.43 Axiom 2 (mZeroNum): aElementOf0(sz00, szNzAzT0) = true2.
% 8.07/1.43 Axiom 3 (m__3623_1): sdtlpdtrp0(xN, sz00) = xS.
% 8.07/1.43 Axiom 4 (mZeroLess): fresh69(X, X, Y) = true2.
% 8.07/1.43 Axiom 5 (m__3754_1): fresh172(X, X, Y, Z) = true2.
% 8.07/1.43 Axiom 6 (mZeroLess): fresh69(aElementOf0(X, szNzAzT0), true2, X) = sdtlseqdt0(sz00, X).
% 8.07/1.43 Axiom 7 (m__3754_1): fresh17(X, X, Y, Z) = aSubsetOf0(sdtlpdtrp0(xN, Y), sdtlpdtrp0(xN, Z)).
% 8.07/1.43 Axiom 8 (m__3754_1): fresh171(X, X, Y, Z) = fresh172(aElementOf0(Y, szNzAzT0), true2, Y, Z).
% 8.07/1.43 Axiom 9 (m__3754_1): fresh171(sdtlseqdt0(X, Y), true2, Y, X) = fresh17(aElementOf0(X, szNzAzT0), true2, Y, X).
% 8.07/1.43
% 8.07/1.43 Goal 1 (m___2): aSubsetOf0(sdtlpdtrp0(xN, xi), xS) = true2.
% 8.07/1.43 Proof:
% 8.07/1.43 aSubsetOf0(sdtlpdtrp0(xN, xi), xS)
% 8.07/1.43 = { by axiom 3 (m__3623_1) R->L }
% 8.07/1.43 aSubsetOf0(sdtlpdtrp0(xN, xi), sdtlpdtrp0(xN, sz00))
% 8.07/1.43 = { by axiom 7 (m__3754_1) R->L }
% 8.07/1.43 fresh17(true2, true2, xi, sz00)
% 8.07/1.43 = { by axiom 2 (mZeroNum) R->L }
% 8.07/1.43 fresh17(aElementOf0(sz00, szNzAzT0), true2, xi, sz00)
% 8.07/1.43 = { by axiom 9 (m__3754_1) R->L }
% 8.07/1.43 fresh171(sdtlseqdt0(sz00, xi), true2, xi, sz00)
% 8.07/1.43 = { by axiom 6 (mZeroLess) R->L }
% 8.07/1.43 fresh171(fresh69(aElementOf0(xi, szNzAzT0), true2, xi), true2, xi, sz00)
% 8.07/1.43 = { by axiom 1 (m__3989) }
% 8.07/1.43 fresh171(fresh69(true2, true2, xi), true2, xi, sz00)
% 8.07/1.43 = { by axiom 4 (mZeroLess) }
% 8.07/1.43 fresh171(true2, true2, xi, sz00)
% 8.07/1.43 = { by axiom 8 (m__3754_1) }
% 8.07/1.43 fresh172(aElementOf0(xi, szNzAzT0), true2, xi, sz00)
% 8.07/1.43 = { by axiom 1 (m__3989) }
% 8.07/1.43 fresh172(true2, true2, xi, sz00)
% 8.07/1.43 = { by axiom 5 (m__3754_1) }
% 8.07/1.43 true2
% 8.07/1.43 % SZS output end Proof
% 8.07/1.43
% 8.07/1.43 RESULT: Theorem (the conjecture is true).
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