TSTP Solution File: NUM548+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM548+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 : n032.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 11:56:54 EDT 2023
% Result : Theorem 0.17s 0.62s
% Output : Proof 0.17s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.10 % Problem : NUM548+3 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.11 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.31 % Computer : n032.cluster.edu
% 0.14/0.31 % Model : x86_64 x86_64
% 0.14/0.31 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.31 % Memory : 8042.1875MB
% 0.14/0.31 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.31 % CPULimit : 300
% 0.14/0.31 % WCLimit : 300
% 0.14/0.31 % DateTime : Fri Aug 25 14:31:57 EDT 2023
% 0.14/0.31 % CPUTime :
% 0.17/0.62 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.17/0.62
% 0.17/0.62 % SZS status Theorem
% 0.17/0.62
% 0.17/0.62 % SZS output start Proof
% 0.17/0.62 Take the following subset of the input axioms:
% 0.17/0.62 fof(mCardNum, axiom, ![W0]: (aSet0(W0) => (aElementOf0(sbrdtbr0(W0), szNzAzT0) <=> isFinite0(W0)))).
% 0.17/0.62 fof(m__, conjecture, isFinite0(xQ)).
% 0.17/0.62 fof(m__2202, hypothesis, aElementOf0(xk, szNzAzT0)).
% 0.17/0.62 fof(m__2270, hypothesis, aSet0(xQ) & (![W0_2]: (aElementOf0(W0_2, xQ) => aElementOf0(W0_2, xS)) & (aSubsetOf0(xQ, xS) & (sbrdtbr0(xQ)=xk & aElementOf0(xQ, slbdtsldtrb0(xS, xk)))))).
% 0.17/0.62
% 0.17/0.62 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.17/0.62 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.17/0.62 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.17/0.62 fresh(y, y, x1...xn) = u
% 0.17/0.62 C => fresh(s, t, x1...xn) = v
% 0.17/0.62 where fresh is a fresh function symbol and x1..xn are the free
% 0.17/0.62 variables of u and v.
% 0.17/0.62 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.17/0.62 input problem has no model of domain size 1).
% 0.17/0.62
% 0.17/0.62 The encoding turns the above axioms into the following unit equations and goals:
% 0.17/0.62
% 0.17/0.62 Axiom 1 (m__2270_1): aSet0(xQ) = true2.
% 0.17/0.62 Axiom 2 (m__2270): sbrdtbr0(xQ) = xk.
% 0.17/0.62 Axiom 3 (m__2202): aElementOf0(xk, szNzAzT0) = true2.
% 0.17/0.62 Axiom 4 (mCardNum): fresh89(X, X, Y) = isFinite0(Y).
% 0.17/0.62 Axiom 5 (mCardNum): fresh88(X, X, Y) = true2.
% 0.17/0.62 Axiom 6 (mCardNum): fresh89(aElementOf0(sbrdtbr0(X), szNzAzT0), true2, X) = fresh88(aSet0(X), true2, X).
% 0.17/0.62
% 0.17/0.62 Goal 1 (m__): isFinite0(xQ) = true2.
% 0.17/0.62 Proof:
% 0.17/0.62 isFinite0(xQ)
% 0.17/0.62 = { by axiom 4 (mCardNum) R->L }
% 0.17/0.62 fresh89(true2, true2, xQ)
% 0.17/0.62 = { by axiom 3 (m__2202) R->L }
% 0.17/0.62 fresh89(aElementOf0(xk, szNzAzT0), true2, xQ)
% 0.17/0.62 = { by axiom 2 (m__2270) R->L }
% 0.17/0.62 fresh89(aElementOf0(sbrdtbr0(xQ), szNzAzT0), true2, xQ)
% 0.17/0.62 = { by axiom 6 (mCardNum) }
% 0.17/0.62 fresh88(aSet0(xQ), true2, xQ)
% 0.17/0.62 = { by axiom 1 (m__2270_1) }
% 0.17/0.62 fresh88(true2, true2, xQ)
% 0.17/0.62 = { by axiom 5 (mCardNum) }
% 0.17/0.62 true2
% 0.17/0.62 % SZS output end Proof
% 0.17/0.62
% 0.17/0.62 RESULT: Theorem (the conjecture is true).
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