TSTP Solution File: NUM560+2 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM560+2 : 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 : n031.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 11:56:59 EDT 2023
% Result : Theorem 0.20s 0.74s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.13 % Problem : NUM560+2 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35 % Computer : n031.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Fri Aug 25 09:31:36 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.20/0.74 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.74
% 0.20/0.74 % SZS status Theorem
% 0.20/0.74
% 0.20/0.74 % SZS output start Proof
% 0.20/0.74 Take the following subset of the input axioms:
% 0.20/0.74 fof(m__, conjecture, (aSet0(sdtlbdtrb0(xF, xy)) & ![W0]: (aElementOf0(W0, sdtlbdtrb0(xF, xy)) <=> (aElementOf0(W0, szDzozmdt0(xF)) & sdtlpdtrp0(xF, W0)=xy))) => (![W0_2]: (aElementOf0(W0_2, sdtlbdtrb0(xF, xy)) => aElementOf0(W0_2, szDzozmdt0(xF))) | aSubsetOf0(sdtlbdtrb0(xF, xy), szDzozmdt0(xF)))).
% 0.20/0.74
% 0.20/0.74 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.74 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.74 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.74 fresh(y, y, x1...xn) = u
% 0.20/0.74 C => fresh(s, t, x1...xn) = v
% 0.20/0.74 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.74 variables of u and v.
% 0.20/0.74 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.74 input problem has no model of domain size 1).
% 0.20/0.74
% 0.20/0.74 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.74
% 0.20/0.74 Axiom 1 (m___1): aElementOf0(w0, sdtlbdtrb0(xF, xy)) = true2.
% 0.20/0.74 Axiom 2 (m___4): fresh7(X, X, Y) = true2.
% 0.20/0.74 Axiom 3 (m___4): fresh7(aElementOf0(X, sdtlbdtrb0(xF, xy)), true2, X) = aElementOf0(X, szDzozmdt0(xF)).
% 0.20/0.74
% 0.20/0.74 Goal 1 (m___5): aElementOf0(w0, szDzozmdt0(xF)) = true2.
% 0.20/0.74 Proof:
% 0.20/0.74 aElementOf0(w0, szDzozmdt0(xF))
% 0.20/0.74 = { by axiom 3 (m___4) R->L }
% 0.20/0.74 fresh7(aElementOf0(w0, sdtlbdtrb0(xF, xy)), true2, w0)
% 0.20/0.74 = { by axiom 1 (m___1) }
% 0.20/0.74 fresh7(true2, true2, w0)
% 0.20/0.74 = { by axiom 2 (m___4) }
% 0.20/0.74 true2
% 0.20/0.74 % SZS output end Proof
% 0.20/0.74
% 0.20/0.74 RESULT: Theorem (the conjecture is true).
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