TSTP Solution File: NUM557+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM557+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 : n018.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 11:56:57 EDT 2023
% Result : Theorem 0.19s 0.75s
% Output : Proof 0.19s
% 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 : NUM557+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
% 0.13/0.34 % Computer : n018.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 15:22:28 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.75 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.75
% 0.19/0.75 % SZS status Theorem
% 0.19/0.75
% 0.19/0.75 % SZS output start Proof
% 0.19/0.75 Take the following subset of the input axioms:
% 0.19/0.75 fof(m__, conjecture, ((![W0]: (aElementOf0(W0, xP) => aElementOf0(W0, xS)) | aSubsetOf0(xP, xS)) & sbrdtbr0(xP)=xk) | aElementOf0(xP, slbdtsldtrb0(xS, xk))).
% 0.19/0.75 fof(m__2431, hypothesis, ![W0_2]: (aElementOf0(W0_2, xP) => aElementOf0(W0_2, xS)) & (aSubsetOf0(xP, xS) & sbrdtbr0(xP)=xk)).
% 0.19/0.75
% 0.19/0.75 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.75 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.75 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.75 fresh(y, y, x1...xn) = u
% 0.19/0.75 C => fresh(s, t, x1...xn) = v
% 0.19/0.75 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.75 variables of u and v.
% 0.19/0.75 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.75 input problem has no model of domain size 1).
% 0.19/0.75
% 0.19/0.75 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.75
% 0.19/0.75 Axiom 1 (m__2431): sbrdtbr0(xP) = xk.
% 0.19/0.75 Axiom 2 (m__2431_1): aSubsetOf0(xP, xS) = true2.
% 0.19/0.75
% 0.19/0.75 Goal 1 (m___2): tuple4(sbrdtbr0(xP), aSubsetOf0(xP, xS)) = tuple4(xk, true2).
% 0.19/0.75 Proof:
% 0.19/0.75 tuple4(sbrdtbr0(xP), aSubsetOf0(xP, xS))
% 0.19/0.75 = { by axiom 1 (m__2431) }
% 0.19/0.75 tuple4(xk, aSubsetOf0(xP, xS))
% 0.19/0.75 = { by axiom 2 (m__2431_1) }
% 0.19/0.75 tuple4(xk, true2)
% 0.19/0.75 % SZS output end Proof
% 0.19/0.75
% 0.19/0.75 RESULT: Theorem (the conjecture is true).
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