TSTP Solution File: NUM540+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM540+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 11:56:51 EDT 2023
% Result : Theorem 0.20s 0.57s
% 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
% 0.00/0.12 % Problem : NUM540+2 : 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
% 0.14/0.34 % Computer : n031.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Fri Aug 25 16:53:22 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.20/0.57 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.20/0.57
% 0.20/0.57 % SZS status Theorem
% 0.20/0.57
% 0.20/0.58 % SZS output start Proof
% 0.20/0.58 Take the following subset of the input axioms:
% 0.20/0.58 fof(mCountNFin, axiom, ![W0]: ((aSet0(W0) & isCountable0(W0)) => ~isFinite0(W0))).
% 0.20/0.58 fof(mCountNFin_01, axiom, ![W0_2]: ((aSet0(W0_2) & isCountable0(W0_2)) => W0_2!=slcrc0)).
% 0.20/0.58 fof(mDefDiff, definition, ![W1, W0_2]: ((aSet0(W0_2) & aElement0(W1)) => ![W2]: (W2=sdtmndt0(W0_2, W1) <=> (aSet0(W2) & ![W3]: (aElementOf0(W3, W2) <=> (aElement0(W3) & (aElementOf0(W3, W0_2) & W3!=W1))))))).
% 0.20/0.58 fof(mDefEmp, definition, ![W0_2]: (W0_2=slcrc0 <=> (aSet0(W0_2) & ~?[W1_2]: aElementOf0(W1_2, W0_2)))).
% 0.20/0.58 fof(mNatNSucc, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => W0_2!=szszuzczcdt0(W0_2))).
% 0.20/0.58 fof(mNoScLessZr, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ~sdtlseqdt0(szszuzczcdt0(W0_2), sz00))).
% 0.20/0.58 fof(mSuccNum, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => (aElementOf0(szszuzczcdt0(W0_2), szNzAzT0) & szszuzczcdt0(W0_2)!=sz00))).
% 0.20/0.58 fof(m__, conjecture, (aSet0(slbdtrb0(sz00)) & ![W0_2]: ~aElementOf0(W0_2, slbdtrb0(sz00))) => (~?[W0_2]: aElementOf0(W0_2, slbdtrb0(sz00)) | slbdtrb0(sz00)=slcrc0)).
% 0.20/0.58
% 0.20/0.58 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.58 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.58 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.58 fresh(y, y, x1...xn) = u
% 0.20/0.58 C => fresh(s, t, x1...xn) = v
% 0.20/0.58 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.58 variables of u and v.
% 0.20/0.58 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.58 input problem has no model of domain size 1).
% 0.20/0.58
% 0.20/0.58 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.58
% 0.20/0.58 Axiom 1 (m___1): aElementOf0(w0, slbdtrb0(sz00)) = true2.
% 0.20/0.58
% 0.20/0.58 Goal 1 (m___3): aElementOf0(X, slbdtrb0(sz00)) = true2.
% 0.20/0.58 The goal is true when:
% 0.20/0.58 X = w0
% 0.20/0.58
% 0.20/0.58 Proof:
% 0.20/0.58 aElementOf0(w0, slbdtrb0(sz00))
% 0.20/0.58 = { by axiom 1 (m___1) }
% 0.20/0.58 true2
% 0.20/0.58 % SZS output end Proof
% 0.20/0.58
% 0.20/0.58 RESULT: Theorem (the conjecture is true).
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