TSTP Solution File: NUM584+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM584+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 : n025.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 11:57:07 EDT 2023
% Result : Theorem 5.18s 1.06s
% Output : Proof 5.18s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM584+3 : 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
% 0.13/0.33 % Computer : n025.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Fri Aug 25 08:02:37 EDT 2023
% 0.13/0.34 % CPUTime :
% 5.18/1.06 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 5.18/1.06
% 5.18/1.06 % SZS status Theorem
% 5.18/1.06
% 5.18/1.06 % SZS output start Proof
% 5.18/1.06 Take the following subset of the input axioms:
% 5.18/1.07 fof(m__, conjecture, (aElementOf0(szmzizndt0(sdtlpdtrp0(xN, xi)), sdtlpdtrp0(xN, xi)) & ![W0]: (aElementOf0(W0, sdtlpdtrp0(xN, xi)) => sdtlseqdt0(szmzizndt0(sdtlpdtrp0(xN, xi)), W0))) => ((aSet0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) & ![W0_2]: (aElementOf0(W0_2, sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) <=> (aElement0(W0_2) & (aElementOf0(W0_2, xQ) | W0_2=szmzizndt0(sdtlpdtrp0(xN, xi)))))) => (((![W0_2]: (aElementOf0(W0_2, sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) => aElementOf0(W0_2, xS)) | aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS)) & sbrdtbr0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))))=xK) | aElementOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), slbdtsldtrb0(xS, xK))))).
% 5.18/1.07 fof(m__4007, hypothesis, aElementOf0(szmzizndt0(sdtlpdtrp0(xN, xi)), sdtlpdtrp0(xN, xi)) & (![W0_2]: (aElementOf0(W0_2, sdtlpdtrp0(xN, xi)) => sdtlseqdt0(szmzizndt0(sdtlpdtrp0(xN, xi)), W0_2)) & (aSet0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) & (![W0_2]: (aElementOf0(W0_2, sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) <=> (aElement0(W0_2) & (aElementOf0(W0_2, xQ) | W0_2=szmzizndt0(sdtlpdtrp0(xN, xi))))) & sbrdtbr0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))))=xK)))).
% 5.18/1.07 fof(m__4024, hypothesis, aElementOf0(szmzizndt0(sdtlpdtrp0(xN, xi)), sdtlpdtrp0(xN, xi)) & (![W0_2]: (aElementOf0(W0_2, sdtlpdtrp0(xN, xi)) => sdtlseqdt0(szmzizndt0(sdtlpdtrp0(xN, xi)), W0_2)) & (aSet0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) & (![W0_2]: (aElementOf0(W0_2, sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) <=> (aElement0(W0_2) & (aElementOf0(W0_2, xQ) | W0_2=szmzizndt0(sdtlpdtrp0(xN, xi))))) & (![W0_2]: (aElementOf0(W0_2, sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) => aElementOf0(W0_2, xS)) & aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS)))))).
% 5.18/1.07
% 5.18/1.07 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.18/1.07 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.18/1.07 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.18/1.07 fresh(y, y, x1...xn) = u
% 5.18/1.07 C => fresh(s, t, x1...xn) = v
% 5.18/1.07 where fresh is a fresh function symbol and x1..xn are the free
% 5.18/1.07 variables of u and v.
% 5.18/1.07 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.18/1.07 input problem has no model of domain size 1).
% 5.18/1.07
% 5.18/1.07 The encoding turns the above axioms into the following unit equations and goals:
% 5.18/1.07
% 5.18/1.07 Axiom 1 (m__4007): sbrdtbr0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))) = xK.
% 5.18/1.07 Axiom 2 (m__4024_2): aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS) = true2.
% 5.18/1.07
% 5.18/1.07 Goal 1 (m___5): tuple4(sbrdtbr0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))), aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS)) = tuple4(xK, true2).
% 5.18/1.07 Proof:
% 5.18/1.07 tuple4(sbrdtbr0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi)))), aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS))
% 5.18/1.07 = { by axiom 1 (m__4007) }
% 5.18/1.07 tuple4(xK, aSubsetOf0(sdtpldt0(xQ, szmzizndt0(sdtlpdtrp0(xN, xi))), xS))
% 5.18/1.07 = { by axiom 2 (m__4024_2) }
% 5.18/1.07 tuple4(xK, true2)
% 5.18/1.07 % SZS output end Proof
% 5.18/1.07
% 5.18/1.07 RESULT: Theorem (the conjecture is true).
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