TSTP Solution File: RNG105+2 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : RNG105+2 : 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 : n022.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 13:59:19 EDT 2023
% Result : Theorem 5.72s 1.11s
% Output : Proof 5.72s
% 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 : RNG105+2 : 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 : n022.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 : Sun Aug 27 02:44:40 EDT 2023
% 0.13/0.35 % CPUTime :
% 5.72/1.11 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 5.72/1.11
% 5.72/1.11 % SZS status Theorem
% 5.72/1.11
% 5.72/1.11 % SZS output start Proof
% 5.72/1.11 Take the following subset of the input axioms:
% 5.72/1.11 fof(mSortsB, axiom, ![W0, W1]: ((aElement0(W0) & aElement0(W1)) => aElement0(sdtpldt0(W0, W1)))).
% 5.72/1.11 fof(mSortsB_02, axiom, ![W0_3, W1_2]: ((aElement0(W0_3) & aElement0(W1_2)) => aElement0(sdtasdt0(W0_3, W1_2)))).
% 5.72/1.11 fof(m__, conjecture, (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xc, W0_2)=sdtpldt0(xx, xy)) | aElementOf0(sdtpldt0(xx, xy), slsdtgt0(xc))) & (?[W0_3]: (aElement0(W0_3) & sdtasdt0(xc, W0_3)=sdtasdt0(xz, xx)) | aElementOf0(sdtasdt0(xz, xx), slsdtgt0(xc)))).
% 5.72/1.11 fof(m__1933, hypothesis, ?[W0_3]: (aElement0(W0_3) & sdtasdt0(xc, W0_3)=xx) & (aElementOf0(xx, slsdtgt0(xc)) & (?[W0_3]: (aElement0(W0_3) & sdtasdt0(xc, W0_3)=xy) & (aElementOf0(xy, slsdtgt0(xc)) & aElement0(xz))))).
% 5.72/1.11 fof(m__1956, hypothesis, aElement0(xu) & sdtasdt0(xc, xu)=xx).
% 5.72/1.11 fof(m__1979, hypothesis, aElement0(xv) & sdtasdt0(xc, xv)=xy).
% 5.72/1.11 fof(m__2010, hypothesis, sdtpldt0(xx, xy)=sdtasdt0(xc, sdtpldt0(xu, xv))).
% 5.72/1.11 fof(m__2043, hypothesis, sdtasdt0(xz, xx)=sdtasdt0(xc, sdtasdt0(xu, xz))).
% 5.72/1.11
% 5.72/1.11 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.72/1.11 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.72/1.11 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.72/1.11 fresh(y, y, x1...xn) = u
% 5.72/1.11 C => fresh(s, t, x1...xn) = v
% 5.72/1.11 where fresh is a fresh function symbol and x1..xn are the free
% 5.72/1.11 variables of u and v.
% 5.72/1.11 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.72/1.11 input problem has no model of domain size 1).
% 5.72/1.11
% 5.72/1.11 The encoding turns the above axioms into the following unit equations and goals:
% 5.72/1.11
% 5.72/1.11 Axiom 1 (m__1933_2): aElement0(xz) = true2.
% 5.72/1.11 Axiom 2 (m__1956_1): aElement0(xu) = true2.
% 5.72/1.11 Axiom 3 (m__1979_1): aElement0(xv) = true2.
% 5.72/1.11 Axiom 4 (m__2043): sdtasdt0(xz, xx) = sdtasdt0(xc, sdtasdt0(xu, xz)).
% 5.72/1.11 Axiom 5 (m__2010): sdtpldt0(xx, xy) = sdtasdt0(xc, sdtpldt0(xu, xv)).
% 5.72/1.11 Axiom 6 (mSortsB): fresh11(X, X, Y, Z) = aElement0(sdtpldt0(Y, Z)).
% 5.72/1.11 Axiom 7 (mSortsB): fresh10(X, X, Y, Z) = true2.
% 5.72/1.11 Axiom 8 (mSortsB_02): fresh9(X, X, Y, Z) = aElement0(sdtasdt0(Y, Z)).
% 5.72/1.11 Axiom 9 (mSortsB_02): fresh8(X, X, Y, Z) = true2.
% 5.72/1.11 Axiom 10 (mSortsB): fresh11(aElement0(X), true2, Y, X) = fresh10(aElement0(Y), true2, Y, X).
% 5.72/1.11 Axiom 11 (mSortsB_02): fresh9(aElement0(X), true2, Y, X) = fresh8(aElement0(Y), true2, Y, X).
% 5.72/1.11
% 5.72/1.11 Goal 1 (m__): tuple3(sdtasdt0(xc, X), sdtasdt0(xc, Y), aElement0(X), aElement0(Y)) = tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), true2, true2).
% 5.72/1.11 The goal is true when:
% 5.72/1.11 X = sdtpldt0(xu, xv)
% 5.72/1.11 Y = sdtasdt0(xu, xz)
% 5.72/1.11
% 5.72/1.11 Proof:
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xc, sdtasdt0(xu, xz)), aElement0(sdtpldt0(xu, xv)), aElement0(sdtasdt0(xu, xz)))
% 5.72/1.11 = { by axiom 4 (m__2043) R->L }
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), aElement0(sdtasdt0(xu, xz)))
% 5.72/1.11 = { by axiom 8 (mSortsB_02) R->L }
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), fresh9(true2, true2, xu, xz))
% 5.72/1.11 = { by axiom 1 (m__1933_2) R->L }
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), fresh9(aElement0(xz), true2, xu, xz))
% 5.72/1.11 = { by axiom 11 (mSortsB_02) }
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), fresh8(aElement0(xu), true2, xu, xz))
% 5.72/1.11 = { by axiom 2 (m__1956_1) }
% 5.72/1.11 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), fresh8(true2, true2, xu, xz))
% 5.72/1.12 = { by axiom 9 (mSortsB_02) }
% 5.72/1.12 tuple3(sdtasdt0(xc, sdtpldt0(xu, xv)), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), true2)
% 5.72/1.12 = { by axiom 5 (m__2010) R->L }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), aElement0(sdtpldt0(xu, xv)), true2)
% 5.72/1.12 = { by axiom 6 (mSortsB) R->L }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), fresh11(true2, true2, xu, xv), true2)
% 5.72/1.12 = { by axiom 3 (m__1979_1) R->L }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), fresh11(aElement0(xv), true2, xu, xv), true2)
% 5.72/1.12 = { by axiom 10 (mSortsB) }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), fresh10(aElement0(xu), true2, xu, xv), true2)
% 5.72/1.12 = { by axiom 2 (m__1956_1) }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), fresh10(true2, true2, xu, xv), true2)
% 5.72/1.12 = { by axiom 7 (mSortsB) }
% 5.72/1.12 tuple3(sdtpldt0(xx, xy), sdtasdt0(xz, xx), true2, true2)
% 5.72/1.12 % SZS output end Proof
% 5.72/1.12
% 5.72/1.12 RESULT: Theorem (the conjecture is true).
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