TSTP Solution File: RNG123+4 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : RNG123+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 : n008.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 13:59:25 EDT 2023
% Result : Theorem 7.47s 1.51s
% Output : Proof 7.47s
% 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 : RNG123+4 : 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
% 0.13/0.33 % Computer : n008.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Sun Aug 27 02:10:02 EDT 2023
% 0.13/0.34 % CPUTime :
% 7.47/1.51 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 7.47/1.51
% 7.47/1.51 % SZS status Theorem
% 7.47/1.51
% 7.47/1.51 % SZS output start Proof
% 7.47/1.51 Take the following subset of the input axioms:
% 7.47/1.52 fof(m__, conjecture, ?[W0, W1]: (aElementOf0(W0, slsdtgt0(xa)) & (aElementOf0(W1, slsdtgt0(xb)) & sdtpldt0(W0, W1)=xr)) | aElementOf0(xr, xI)).
% 7.47/1.52 fof(m__2174, hypothesis, aSet0(xI) & (![W0_2]: (aElementOf0(W0_2, xI) => (![W1_2]: (aElementOf0(W1_2, xI) => aElementOf0(sdtpldt0(W0_2, W1_2), xI)) & ![W1_2]: (aElement0(W1_2) => aElementOf0(sdtasdt0(W1_2, W0_2), xI)))) & (aIdeal0(xI) & (![W0_2]: (aElementOf0(W0_2, slsdtgt0(xa)) <=> ?[W1_2]: (aElement0(W1_2) & sdtasdt0(xa, W1_2)=W0_2)) & (![W0_2]: (aElementOf0(W0_2, slsdtgt0(xb)) <=> ?[W1_2]: (aElement0(W1_2) & sdtasdt0(xb, W1_2)=W0_2)) & (![W0_2]: (aElementOf0(W0_2, xI) <=> ?[W2, W1_2]: (aElementOf0(W1_2, slsdtgt0(xa)) & (aElementOf0(W2, slsdtgt0(xb)) & sdtpldt0(W1_2, W2)=W0_2))) & xI=sdtpldt1(slsdtgt0(xa), slsdtgt0(xb)))))))).
% 7.47/1.52 fof(m__2690, hypothesis, ?[W0_2, W1_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1_2, slsdtgt0(xb)) & sdtpldt0(W0_2, W1_2)=smndt0(sdtasdt0(xq, xu)))) & aElementOf0(smndt0(sdtasdt0(xq, xu)), xI)).
% 7.47/1.52 fof(m__2699, hypothesis, ?[W0_2, W1_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1_2, slsdtgt0(xb)) & sdtpldt0(W0_2, W1_2)=xb)) & (?[W0_2, W1_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1_2, slsdtgt0(xb)) & sdtpldt0(W0_2, W1_2)=xb)) & aElementOf0(xb, xI))).
% 7.47/1.52 fof(m__2718, hypothesis, xr=sdtpldt0(smndt0(sdtasdt0(xq, xu)), xb)).
% 7.47/1.52
% 7.47/1.52 Now clausify the problem and encode Horn clauses using encoding 3 of
% 7.47/1.52 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 7.47/1.52 We repeatedly replace C & s=t => u=v by the two clauses:
% 7.47/1.52 fresh(y, y, x1...xn) = u
% 7.47/1.52 C => fresh(s, t, x1...xn) = v
% 7.47/1.52 where fresh is a fresh function symbol and x1..xn are the free
% 7.47/1.52 variables of u and v.
% 7.47/1.52 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 7.47/1.52 input problem has no model of domain size 1).
% 7.47/1.52
% 7.47/1.52 The encoding turns the above axioms into the following unit equations and goals:
% 7.47/1.52
% 7.47/1.52 Axiom 1 (m__2699_2): aElementOf0(xb, xI) = true2.
% 7.47/1.52 Axiom 2 (m__2690): sdtpldt0(w0_3, w1_3) = smndt0(sdtasdt0(xq, xu)).
% 7.47/1.52 Axiom 3 (m__2174_7): fresh26(X, X, Y, Z) = aElementOf0(sdtpldt0(Y, Z), xI).
% 7.47/1.52 Axiom 4 (m__2174_7): fresh25(X, X, Y, Z) = true2.
% 7.47/1.52 Axiom 5 (m__2690_1): aElementOf0(smndt0(sdtasdt0(xq, xu)), xI) = true2.
% 7.47/1.52 Axiom 6 (m__2718): xr = sdtpldt0(smndt0(sdtasdt0(xq, xu)), xb).
% 7.47/1.52 Axiom 7 (m__2174_7): fresh26(aElementOf0(X, xI), true2, Y, X) = fresh25(aElementOf0(Y, xI), true2, Y, X).
% 7.47/1.52
% 7.47/1.52 Goal 1 (m___1): aElementOf0(xr, xI) = true2.
% 7.47/1.52 Proof:
% 7.47/1.52 aElementOf0(xr, xI)
% 7.47/1.52 = { by axiom 6 (m__2718) }
% 7.47/1.52 aElementOf0(sdtpldt0(smndt0(sdtasdt0(xq, xu)), xb), xI)
% 7.47/1.52 = { by axiom 2 (m__2690) R->L }
% 7.47/1.52 aElementOf0(sdtpldt0(sdtpldt0(w0_3, w1_3), xb), xI)
% 7.47/1.52 = { by axiom 3 (m__2174_7) R->L }
% 7.47/1.52 fresh26(true2, true2, sdtpldt0(w0_3, w1_3), xb)
% 7.47/1.52 = { by axiom 1 (m__2699_2) R->L }
% 7.47/1.52 fresh26(aElementOf0(xb, xI), true2, sdtpldt0(w0_3, w1_3), xb)
% 7.47/1.52 = { by axiom 7 (m__2174_7) }
% 7.47/1.52 fresh25(aElementOf0(sdtpldt0(w0_3, w1_3), xI), true2, sdtpldt0(w0_3, w1_3), xb)
% 7.47/1.52 = { by axiom 2 (m__2690) }
% 7.47/1.52 fresh25(aElementOf0(smndt0(sdtasdt0(xq, xu)), xI), true2, sdtpldt0(w0_3, w1_3), xb)
% 7.47/1.52 = { by axiom 5 (m__2690_1) }
% 7.47/1.52 fresh25(true2, true2, sdtpldt0(w0_3, w1_3), xb)
% 7.47/1.52 = { by axiom 4 (m__2174_7) }
% 7.47/1.52 true2
% 7.47/1.52 % SZS output end Proof
% 7.47/1.52
% 7.47/1.52 RESULT: Theorem (the conjecture is true).
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