TSTP Solution File: RNG102+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : RNG102+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 : n001.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 13:59:17 EDT 2023
% Result : Theorem 109.60s 14.56s
% Output : Proof 109.60s
% 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 : RNG102+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
% 0.13/0.34 % Computer : n001.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % 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 02:44:32 EDT 2023
% 0.13/0.34 % CPUTime :
% 109.60/14.56 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 109.60/14.56
% 109.60/14.56 % SZS status Theorem
% 109.60/14.56
% 109.60/14.57 % SZS output start Proof
% 109.60/14.57 Take the following subset of the input axioms:
% 109.60/14.57 fof(mDefPrIdeal, definition, ![W0]: (aElement0(W0) => ![W1]: (W1=slsdtgt0(W0) <=> (aSet0(W1) & ![W2]: (aElementOf0(W2, W1) <=> ?[W3]: (aElement0(W3) & sdtasdt0(W0, W3)=W2)))))).
% 109.60/14.57 fof(m__, conjecture, ?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=xy)).
% 109.60/14.57 fof(m__1905, hypothesis, aElement0(xc)).
% 109.60/14.57 fof(m__1933, hypothesis, aElementOf0(xx, slsdtgt0(xc)) & (aElementOf0(xy, slsdtgt0(xc)) & aElement0(xz))).
% 109.60/14.57
% 109.60/14.57 Now clausify the problem and encode Horn clauses using encoding 3 of
% 109.60/14.57 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 109.60/14.57 We repeatedly replace C & s=t => u=v by the two clauses:
% 109.60/14.57 fresh(y, y, x1...xn) = u
% 109.60/14.57 C => fresh(s, t, x1...xn) = v
% 109.60/14.57 where fresh is a fresh function symbol and x1..xn are the free
% 109.60/14.57 variables of u and v.
% 109.60/14.57 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 109.60/14.57 input problem has no model of domain size 1).
% 109.60/14.57
% 109.60/14.57 The encoding turns the above axioms into the following unit equations and goals:
% 109.60/14.57
% 109.60/14.57 Axiom 1 (m__1905): aElement0(xc) = true2.
% 109.60/14.57 Axiom 2 (m__1933_2): aElementOf0(xy, slsdtgt0(xc)) = true2.
% 109.60/14.57 Axiom 3 (mDefPrIdeal_6): fresh(X, X, Y, Z) = Z.
% 109.60/14.57 Axiom 4 (mDefPrIdeal_1): fresh64(X, X, Y, Z) = true2.
% 109.60/14.57 Axiom 5 (mDefPrIdeal_7): fresh36(X, X, Y, Z) = true2.
% 109.60/14.57 Axiom 6 (mDefPrIdeal_1): fresh63(X, X, Y, Z, W) = fresh64(Z, slsdtgt0(Y), Y, W).
% 109.60/14.57 Axiom 7 (mDefPrIdeal_1): fresh40(X, X, Y, Z, W) = equiv(Y, W).
% 109.60/14.57 Axiom 8 (mDefPrIdeal_6): fresh(equiv(X, Y), true2, X, Y) = sdtasdt0(X, w3(X, Y)).
% 109.60/14.57 Axiom 9 (mDefPrIdeal_7): fresh36(equiv(X, Y), true2, X, Y) = aElement0(w3(X, Y)).
% 109.60/14.57 Axiom 10 (mDefPrIdeal_1): fresh63(aElementOf0(X, Y), true2, Z, Y, X) = fresh40(aElement0(Z), true2, Z, Y, X).
% 109.60/14.57
% 109.60/14.57 Lemma 11: equiv(xc, xy) = true2.
% 109.60/14.57 Proof:
% 109.60/14.57 equiv(xc, xy)
% 109.60/14.57 = { by axiom 7 (mDefPrIdeal_1) R->L }
% 109.60/14.57 fresh40(true2, true2, xc, slsdtgt0(xc), xy)
% 109.60/14.57 = { by axiom 1 (m__1905) R->L }
% 109.60/14.57 fresh40(aElement0(xc), true2, xc, slsdtgt0(xc), xy)
% 109.60/14.57 = { by axiom 10 (mDefPrIdeal_1) R->L }
% 109.60/14.57 fresh63(aElementOf0(xy, slsdtgt0(xc)), true2, xc, slsdtgt0(xc), xy)
% 109.60/14.57 = { by axiom 2 (m__1933_2) }
% 109.60/14.57 fresh63(true2, true2, xc, slsdtgt0(xc), xy)
% 109.60/14.57 = { by axiom 6 (mDefPrIdeal_1) }
% 109.60/14.57 fresh64(slsdtgt0(xc), slsdtgt0(xc), xc, xy)
% 109.60/14.57 = { by axiom 4 (mDefPrIdeal_1) }
% 109.60/14.57 true2
% 109.60/14.57
% 109.60/14.57 Goal 1 (m__): tuple(sdtasdt0(xc, X), aElement0(X)) = tuple(xy, true2).
% 109.60/14.57 The goal is true when:
% 109.60/14.57 X = w3(xc, xy)
% 109.60/14.57
% 109.60/14.57 Proof:
% 109.60/14.57 tuple(sdtasdt0(xc, w3(xc, xy)), aElement0(w3(xc, xy)))
% 109.60/14.57 = { by axiom 8 (mDefPrIdeal_6) R->L }
% 109.60/14.57 tuple(fresh(equiv(xc, xy), true2, xc, xy), aElement0(w3(xc, xy)))
% 109.60/14.57 = { by lemma 11 }
% 109.60/14.57 tuple(fresh(true2, true2, xc, xy), aElement0(w3(xc, xy)))
% 109.60/14.57 = { by axiom 3 (mDefPrIdeal_6) }
% 109.60/14.57 tuple(xy, aElement0(w3(xc, xy)))
% 109.60/14.57 = { by axiom 9 (mDefPrIdeal_7) R->L }
% 109.60/14.57 tuple(xy, fresh36(equiv(xc, xy), true2, xc, xy))
% 109.60/14.57 = { by lemma 11 }
% 109.60/14.57 tuple(xy, fresh36(true2, true2, xc, xy))
% 109.60/14.57 = { by axiom 5 (mDefPrIdeal_7) }
% 109.60/14.57 tuple(xy, true2)
% 109.60/14.57 % SZS output end Proof
% 109.60/14.57
% 109.60/14.57 RESULT: Theorem (the conjecture is true).
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