TSTP Solution File: NUM604+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM604+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 : n002.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 7.61s 1.46s
% Output : Proof 8.35s
% 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 : NUM604+3 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % 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 : n002.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 13:47:32 EDT 2023
% 0.20/0.34 % CPUTime :
% 7.61/1.46 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 7.61/1.46
% 7.61/1.46 % SZS status Theorem
% 7.61/1.46
% 7.61/1.46 % SZS output start Proof
% 7.61/1.46 Take the following subset of the input axioms:
% 8.35/1.46 fof(m__, conjecture, aElementOf0(xx, xS)).
% 8.35/1.46 fof(m__4660, hypothesis, aFunction0(xe) & (szDzozmdt0(xe)=szNzAzT0 & ![W0]: (aElementOf0(W0, szNzAzT0) => (aElementOf0(sdtlpdtrp0(xe, W0), sdtlpdtrp0(xN, W0)) & (![W1]: (aElementOf0(W1, sdtlpdtrp0(xN, W0)) => sdtlseqdt0(sdtlpdtrp0(xe, W0), W1)) & sdtlpdtrp0(xe, W0)=szmzizndt0(sdtlpdtrp0(xN, W0))))))).
% 8.35/1.46 fof(m__5034, hypothesis, aElementOf0(xi, szNzAzT0) & sdtlpdtrp0(xe, xi)=xx).
% 8.35/1.46 fof(m__5045, hypothesis, ![W0_2]: (aElementOf0(W0_2, sdtlpdtrp0(xN, xi)) => aElementOf0(W0_2, xS)) & aSubsetOf0(sdtlpdtrp0(xN, xi), xS)).
% 8.35/1.46
% 8.35/1.46 Now clausify the problem and encode Horn clauses using encoding 3 of
% 8.35/1.46 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 8.35/1.46 We repeatedly replace C & s=t => u=v by the two clauses:
% 8.35/1.46 fresh(y, y, x1...xn) = u
% 8.35/1.46 C => fresh(s, t, x1...xn) = v
% 8.35/1.46 where fresh is a fresh function symbol and x1..xn are the free
% 8.35/1.46 variables of u and v.
% 8.35/1.46 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 8.35/1.46 input problem has no model of domain size 1).
% 8.35/1.46
% 8.35/1.46 The encoding turns the above axioms into the following unit equations and goals:
% 8.35/1.46
% 8.35/1.46 Axiom 1 (m__5034_1): aElementOf0(xi, szNzAzT0) = true2.
% 8.35/1.46 Axiom 2 (m__5034): sdtlpdtrp0(xe, xi) = xx.
% 8.35/1.46 Axiom 3 (m__4660_3): fresh43(X, X, Y) = true2.
% 8.35/1.46 Axiom 4 (m__5045_1): fresh16(X, X, Y) = true2.
% 8.35/1.46 Axiom 5 (m__4660_3): fresh43(aElementOf0(X, szNzAzT0), true2, X) = aElementOf0(sdtlpdtrp0(xe, X), sdtlpdtrp0(xN, X)).
% 8.35/1.46 Axiom 6 (m__5045_1): fresh16(aElementOf0(X, sdtlpdtrp0(xN, xi)), true2, X) = aElementOf0(X, xS).
% 8.35/1.46
% 8.35/1.46 Goal 1 (m__): aElementOf0(xx, xS) = true2.
% 8.35/1.46 Proof:
% 8.35/1.46 aElementOf0(xx, xS)
% 8.35/1.46 = { by axiom 2 (m__5034) R->L }
% 8.35/1.46 aElementOf0(sdtlpdtrp0(xe, xi), xS)
% 8.35/1.46 = { by axiom 6 (m__5045_1) R->L }
% 8.35/1.46 fresh16(aElementOf0(sdtlpdtrp0(xe, xi), sdtlpdtrp0(xN, xi)), true2, sdtlpdtrp0(xe, xi))
% 8.35/1.46 = { by axiom 5 (m__4660_3) R->L }
% 8.35/1.46 fresh16(fresh43(aElementOf0(xi, szNzAzT0), true2, xi), true2, sdtlpdtrp0(xe, xi))
% 8.35/1.46 = { by axiom 1 (m__5034_1) }
% 8.35/1.46 fresh16(fresh43(true2, true2, xi), true2, sdtlpdtrp0(xe, xi))
% 8.35/1.46 = { by axiom 3 (m__4660_3) }
% 8.35/1.46 fresh16(true2, true2, sdtlpdtrp0(xe, xi))
% 8.35/1.46 = { by axiom 4 (m__5045_1) }
% 8.35/1.46 true2
% 8.35/1.46 % SZS output end Proof
% 8.35/1.46
% 8.35/1.46 RESULT: Theorem (the conjecture is true).
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