TSTP Solution File: RNG121+4 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : RNG121+4 : 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 : n005.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 13:59:24 EDT 2023
% Result : Theorem 2.88s 0.79s
% Output : Proof 2.88s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.13 % Problem : RNG121+4 : TPTP v8.1.2. Released v4.0.0.
% 0.08/0.14 % 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 : n005.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 : Sun Aug 27 02:44:38 EDT 2023
% 0.14/0.36 % CPUTime :
% 2.88/0.79 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 2.88/0.79
% 2.88/0.79 % SZS status Theorem
% 2.88/0.79
% 2.88/0.79 % SZS output start Proof
% 2.88/0.79 Take the following subset of the input axioms:
% 2.88/0.80 fof(mAddZero, axiom, ![W0]: (aElement0(W0) => (sdtpldt0(W0, sz00)=W0 & W0=sdtpldt0(sz00, W0)))).
% 2.88/0.80 fof(m__, conjecture, ?[W1, W0_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1, slsdtgt0(xb)) & sdtpldt0(W0_2, W1)=xb)) | (?[W0_2, W1_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1_2, slsdtgt0(xb)) & sdtpldt0(W0_2, W1_2)=xb)) | aElementOf0(xb, xI))).
% 2.88/0.80 fof(m__2091, hypothesis, aElement0(xa) & aElement0(xb)).
% 2.88/0.80 fof(m__2129, hypothesis, aElement0(xc) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=xa) & (doDivides0(xc, xa) & (aDivisorOf0(xc, xa) & (aElement0(xc) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=xb) & (doDivides0(xc, xb) & (aDivisorOf0(xc, xb) & (![W0_2]: ((((aElement0(W0_2) & (?[W1_2]: (aElement0(W1_2) & sdtasdt0(W0_2, W1_2)=xa) | doDivides0(W0_2, xa))) | aDivisorOf0(W0_2, xa)) & (?[W1_2]: (aElement0(W1_2) & sdtasdt0(W0_2, W1_2)=xb) | (doDivides0(W0_2, xb) | aDivisorOf0(W0_2, xb)))) => (?[W1_2]: (aElement0(W1_2) & sdtasdt0(W0_2, W1_2)=xc) & doDivides0(W0_2, xc))) & aGcdOfAnd0(xc, xa, xb)))))))))).
% 2.88/0.80 fof(m__2203, hypothesis, ?[W0_2]: (aElement0(W0_2) & sdtasdt0(xa, W0_2)=sz00) & (aElementOf0(sz00, slsdtgt0(xa)) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xa, W0_2)=xa) & (aElementOf0(xa, slsdtgt0(xa)) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xb, W0_2)=sz00) & (aElementOf0(sz00, slsdtgt0(xb)) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xb, W0_2)=xb) & aElementOf0(xb, slsdtgt0(xb))))))))).
% 2.88/0.80 fof(m__2383, hypothesis, ~((?[W0_2]: (aElement0(W0_2) & sdtasdt0(xu, W0_2)=xa) | (doDivides0(xu, xa) | aDivisorOf0(xu, xa))) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xu, W0_2)=xb) | (doDivides0(xu, xb) | aDivisorOf0(xu, xb))))).
% 2.88/0.80 fof(m__2612, hypothesis, ~(?[W0_2]: (aElement0(W0_2) & sdtasdt0(xu, W0_2)=xb) | doDivides0(xu, xb))).
% 2.88/0.80
% 2.88/0.80 Now clausify the problem and encode Horn clauses using encoding 3 of
% 2.88/0.80 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 2.88/0.80 We repeatedly replace C & s=t => u=v by the two clauses:
% 2.88/0.80 fresh(y, y, x1...xn) = u
% 2.88/0.80 C => fresh(s, t, x1...xn) = v
% 2.88/0.80 where fresh is a fresh function symbol and x1..xn are the free
% 2.88/0.80 variables of u and v.
% 2.88/0.80 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 2.88/0.80 input problem has no model of domain size 1).
% 2.88/0.80
% 2.88/0.80 The encoding turns the above axioms into the following unit equations and goals:
% 2.88/0.80
% 2.88/0.80 Axiom 1 (m__2091_1): aElement0(xb) = true2.
% 2.88/0.80 Axiom 2 (m__2203_11): aElementOf0(xb, slsdtgt0(xb)) = true2.
% 2.88/0.80 Axiom 3 (m__2203_8): aElementOf0(sz00, slsdtgt0(xa)) = true2.
% 2.88/0.80 Axiom 4 (mAddZero_1): fresh12(X, X, Y) = Y.
% 2.88/0.80 Axiom 5 (mAddZero_1): fresh12(aElement0(X), true2, X) = sdtpldt0(sz00, X).
% 2.88/0.80
% 2.88/0.80 Goal 1 (m__): tuple4(sdtpldt0(X, Y), aElementOf0(X, slsdtgt0(xa)), aElementOf0(Y, slsdtgt0(xb))) = tuple4(xb, true2, true2).
% 2.88/0.80 The goal is true when:
% 2.88/0.80 X = sz00
% 2.88/0.80 Y = xb
% 2.88/0.80
% 2.88/0.80 Proof:
% 2.88/0.80 tuple4(sdtpldt0(sz00, xb), aElementOf0(sz00, slsdtgt0(xa)), aElementOf0(xb, slsdtgt0(xb)))
% 2.88/0.80 = { by axiom 5 (mAddZero_1) R->L }
% 2.88/0.80 tuple4(fresh12(aElement0(xb), true2, xb), aElementOf0(sz00, slsdtgt0(xa)), aElementOf0(xb, slsdtgt0(xb)))
% 2.88/0.80 = { by axiom 1 (m__2091_1) }
% 2.88/0.80 tuple4(fresh12(true2, true2, xb), aElementOf0(sz00, slsdtgt0(xa)), aElementOf0(xb, slsdtgt0(xb)))
% 2.88/0.80 = { by axiom 4 (mAddZero_1) }
% 2.88/0.80 tuple4(xb, aElementOf0(sz00, slsdtgt0(xa)), aElementOf0(xb, slsdtgt0(xb)))
% 2.88/0.80 = { by axiom 3 (m__2203_8) }
% 2.88/0.80 tuple4(xb, true2, aElementOf0(xb, slsdtgt0(xb)))
% 2.88/0.80 = { by axiom 2 (m__2203_11) }
% 2.88/0.80 tuple4(xb, true2, true2)
% 2.88/0.80 % SZS output end Proof
% 2.88/0.80
% 2.88/0.80 RESULT: Theorem (the conjecture is true).
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