TSTP Solution File: NUM618+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM618+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 : n011.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 11:57:18 EDT 2023
% Result : Theorem 5.41s 1.08s
% Output : Proof 5.41s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : NUM618+1 : TPTP v8.1.2. Released v4.0.0.
% 0.11/0.13 % 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 : n011.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 12:48:23 EDT 2023
% 0.13/0.35 % CPUTime :
% 5.41/1.08 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 5.41/1.08
% 5.41/1.08 % SZS status Theorem
% 5.41/1.08
% 5.41/1.08 % SZS output start Proof
% 5.41/1.08 Take the following subset of the input axioms:
% 5.41/1.08 fof(mCountNFin, axiom, ![W0]: ((aSet0(W0) & isCountable0(W0)) => ~isFinite0(W0))).
% 5.41/1.08 fof(mCountNFin_01, axiom, ![W0_2]: ((aSet0(W0_2) & isCountable0(W0_2)) => W0_2!=slcrc0)).
% 5.41/1.08 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))))))).
% 5.41/1.08 fof(mDefEmp, definition, ![W0_2]: (W0_2=slcrc0 <=> (aSet0(W0_2) & ~?[W1_2]: aElementOf0(W1_2, W0_2)))).
% 5.41/1.08 fof(mNatNSucc, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => W0_2!=szszuzczcdt0(W0_2))).
% 5.41/1.08 fof(mNoScLessZr, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => ~sdtlseqdt0(szszuzczcdt0(W0_2), sz00))).
% 5.41/1.08 fof(mSuccNum, axiom, ![W0_2]: (aElementOf0(W0_2, szNzAzT0) => (aElementOf0(szszuzczcdt0(W0_2), szNzAzT0) & szszuzczcdt0(W0_2)!=sz00))).
% 5.41/1.08 fof(m__, conjecture, ?[W0_2]: (aElementOf0(W0_2, szNzAzT0) & xx=sdtlpdtrp0(xe, W0_2))).
% 5.41/1.08 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)))).
% 5.41/1.08 fof(m__5365, hypothesis, aElementOf0(xx, szNzAzT0) & aElementOf0(xx, xO)).
% 5.41/1.08
% 5.41/1.08 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.41/1.08 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.41/1.08 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.41/1.08 fresh(y, y, x1...xn) = u
% 5.41/1.08 C => fresh(s, t, x1...xn) = v
% 5.41/1.08 where fresh is a fresh function symbol and x1..xn are the free
% 5.41/1.08 variables of u and v.
% 5.41/1.08 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.41/1.08 input problem has no model of domain size 1).
% 5.41/1.08
% 5.41/1.08 The encoding turns the above axioms into the following unit equations and goals:
% 5.41/1.08
% 5.41/1.08 Axiom 1 (m__5365_1): aElementOf0(xx, xO) = true2.
% 5.41/1.08 Axiom 2 (m__4982): fresh(X, X, Y) = Y.
% 5.41/1.08 Axiom 3 (m__4982_1): fresh11(X, X, Y) = true2.
% 5.41/1.08 Axiom 4 (m__4982): fresh(aElementOf0(X, xO), true2, X) = sdtlpdtrp0(xe, w1(X)).
% 5.41/1.08 Axiom 5 (m__4982_1): fresh11(aElementOf0(X, xO), true2, X) = aElementOf0(w1(X), szNzAzT0).
% 5.41/1.08
% 5.41/1.08 Goal 1 (m__): tuple4(xx, aElementOf0(X, szNzAzT0)) = tuple4(sdtlpdtrp0(xe, X), true2).
% 5.41/1.08 The goal is true when:
% 5.41/1.08 X = w1(xx)
% 5.41/1.08
% 5.41/1.08 Proof:
% 5.41/1.08 tuple4(xx, aElementOf0(w1(xx), szNzAzT0))
% 5.41/1.08 = { by axiom 5 (m__4982_1) R->L }
% 5.41/1.08 tuple4(xx, fresh11(aElementOf0(xx, xO), true2, xx))
% 5.41/1.08 = { by axiom 1 (m__5365_1) }
% 5.41/1.08 tuple4(xx, fresh11(true2, true2, xx))
% 5.41/1.08 = { by axiom 3 (m__4982_1) }
% 5.41/1.08 tuple4(xx, true2)
% 5.41/1.08 = { by axiom 2 (m__4982) R->L }
% 5.41/1.08 tuple4(fresh(true2, true2, xx), true2)
% 5.41/1.08 = { by axiom 1 (m__5365_1) R->L }
% 5.41/1.08 tuple4(fresh(aElementOf0(xx, xO), true2, xx), true2)
% 5.41/1.08 = { by axiom 4 (m__4982) }
% 5.41/1.08 tuple4(sdtlpdtrp0(xe, w1(xx)), true2)
% 5.41/1.08 % SZS output end Proof
% 5.41/1.08
% 5.41/1.08 RESULT: Theorem (the conjecture is true).
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