TSTP Solution File: NUM615+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM615+1 : 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 : n013.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 11:57:17 EDT 2023
% Result : Theorem 4.90s 1.02s
% Output : Proof 4.90s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : NUM615+1 : 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
% 0.13/0.35 % Computer : n013.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Fri Aug 25 11:03:17 EDT 2023
% 0.13/0.35 % CPUTime :
% 4.90/1.02 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 4.90/1.02
% 4.90/1.02 % SZS status Theorem
% 4.90/1.02
% 4.90/1.02 % SZS output start Proof
% 4.90/1.02 Take the following subset of the input axioms:
% 4.90/1.02 fof(mCountNFin, axiom, ![W0]: ((aSet0(W0) & isCountable0(W0)) => ~isFinite0(W0))).
% 4.90/1.02 fof(mCountNFin_01, axiom, ![W0_2]: ((aSet0(W0_2) & isCountable0(W0_2)) => W0_2!=slcrc0)).
% 4.90/1.02 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))))))).
% 4.90/1.02 fof(mDefEmp, definition, ![W0_2]: (W0_2=slcrc0 <=> (aSet0(W0_2) & ~?[W1_2]: aElementOf0(W1_2, W0_2)))).
% 4.90/1.02 fof(mNatNSucc, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => W0_2!=szszuzczcdt0(W0_2))).
% 4.90/1.02 fof(mNoScLessZr, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ~sdtlseqdt0(szszuzczcdt0(W0_2), sz00))).
% 4.90/1.02 fof(mSuccNum, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => (aElementOf0(szszuzczcdt0(W0_2), szNzAzT0) & szszuzczcdt0(W0_2)!=sz00))).
% 4.90/1.02 fof(m__, conjecture, ?[W0_2]: (aElementOf0(W0_2, sdtlbdtrb0(xd, szDzizrdt0(xd))) & sdtlpdtrp0(xe, W0_2)=xp)).
% 4.90/1.02 fof(m__4982, hypothesis, ![W0_2]: (aElementOf0(W0_2, xO) => ?[W1_2]: (aElementOf0(W1_2, szNzAzT0) & (aElementOf0(W1_2, sdtlbdtrb0(xd, szDzizrdt0(xd))) & sdtlpdtrp0(xe, W1_2)=W0_2)))).
% 4.90/1.02 fof(m__5182, hypothesis, aElementOf0(xp, xO)).
% 4.90/1.02
% 4.90/1.02 Now clausify the problem and encode Horn clauses using encoding 3 of
% 4.90/1.02 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 4.90/1.02 We repeatedly replace C & s=t => u=v by the two clauses:
% 4.90/1.02 fresh(y, y, x1...xn) = u
% 4.90/1.02 C => fresh(s, t, x1...xn) = v
% 4.90/1.02 where fresh is a fresh function symbol and x1..xn are the free
% 4.90/1.02 variables of u and v.
% 4.90/1.02 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 4.90/1.02 input problem has no model of domain size 1).
% 4.90/1.02
% 4.90/1.02 The encoding turns the above axioms into the following unit equations and goals:
% 4.90/1.02
% 4.90/1.02 Axiom 1 (m__5182): aElementOf0(xp, xO) = true2.
% 4.90/1.02 Axiom 2 (m__4982): fresh(X, X, Y) = Y.
% 4.90/1.02 Axiom 3 (m__4982_2): fresh10(X, X, Y) = true2.
% 4.90/1.02 Axiom 4 (m__4982): fresh(aElementOf0(X, xO), true2, X) = sdtlpdtrp0(xe, w1(X)).
% 4.90/1.02 Axiom 5 (m__4982_2): fresh10(aElementOf0(X, xO), true2, X) = aElementOf0(w1(X), sdtlbdtrb0(xd, szDzizrdt0(xd))).
% 4.90/1.02
% 4.90/1.02 Goal 1 (m__): tuple4(sdtlpdtrp0(xe, X), aElementOf0(X, sdtlbdtrb0(xd, szDzizrdt0(xd)))) = tuple4(xp, true2).
% 4.90/1.02 The goal is true when:
% 4.90/1.02 X = w1(xp)
% 4.90/1.02
% 4.90/1.02 Proof:
% 4.90/1.02 tuple4(sdtlpdtrp0(xe, w1(xp)), aElementOf0(w1(xp), sdtlbdtrb0(xd, szDzizrdt0(xd))))
% 4.90/1.02 = { by axiom 5 (m__4982_2) R->L }
% 4.90/1.02 tuple4(sdtlpdtrp0(xe, w1(xp)), fresh10(aElementOf0(xp, xO), true2, xp))
% 4.90/1.02 = { by axiom 1 (m__5182) }
% 4.90/1.02 tuple4(sdtlpdtrp0(xe, w1(xp)), fresh10(true2, true2, xp))
% 4.90/1.02 = { by axiom 3 (m__4982_2) }
% 4.90/1.02 tuple4(sdtlpdtrp0(xe, w1(xp)), true2)
% 4.90/1.02 = { by axiom 4 (m__4982) R->L }
% 4.90/1.02 tuple4(fresh(aElementOf0(xp, xO), true2, xp), true2)
% 4.90/1.02 = { by axiom 1 (m__5182) }
% 4.90/1.02 tuple4(fresh(true2, true2, xp), true2)
% 4.90/1.02 = { by axiom 2 (m__4982) }
% 4.90/1.02 tuple4(xp, true2)
% 4.90/1.02 % SZS output end Proof
% 4.90/1.02
% 4.90/1.02 RESULT: Theorem (the conjecture is true).
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