TSTP Solution File: RNG086+2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : RNG086+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 : n028.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 13:59:12 EDT 2023
% Result : Theorem 0.22s 0.52s
% Output : Proof 0.22s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : RNG086+2 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.35 % Computer : n028.cluster.edu
% 0.15/0.35 % Model : x86_64 x86_64
% 0.15/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.35 % Memory : 8042.1875MB
% 0.15/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.35 % CPULimit : 300
% 0.15/0.35 % WCLimit : 300
% 0.15/0.35 % DateTime : Sun Aug 27 02:10:23 EDT 2023
% 0.21/0.35 % CPUTime :
% 0.22/0.52 Command-line arguments: --ground-connectedness --complete-subsets
% 0.22/0.52
% 0.22/0.52 % SZS status Theorem
% 0.22/0.52
% 0.22/0.52 % SZS output start Proof
% 0.22/0.52 Take the following subset of the input axioms:
% 0.22/0.52 fof(m__, conjecture, ?[W0, W1]: (aElementOf0(W0, xI) & (aElementOf0(W1, xJ) & xx=sdtpldt0(W0, W1)))).
% 0.22/0.52 fof(m__901, hypothesis, ?[W0_2, W1_2]: (aElementOf0(W0_2, xI) & (aElementOf0(W1_2, xJ) & sdtpldt0(W0_2, W1_2)=xx)) & (aElementOf0(xx, sdtpldt1(xI, xJ)) & (?[W0_2, W1_2]: (aElementOf0(W0_2, xI) & (aElementOf0(W1_2, xJ) & sdtpldt0(W0_2, W1_2)=xy)) & (aElementOf0(xy, sdtpldt1(xI, xJ)) & aElement0(xz))))).
% 0.22/0.52
% 0.22/0.52 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.22/0.52 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.22/0.52 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.22/0.52 fresh(y, y, x1...xn) = u
% 0.22/0.52 C => fresh(s, t, x1...xn) = v
% 0.22/0.52 where fresh is a fresh function symbol and x1..xn are the free
% 0.22/0.52 variables of u and v.
% 0.22/0.52 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.22/0.52 input problem has no model of domain size 1).
% 0.22/0.52
% 0.22/0.52 The encoding turns the above axioms into the following unit equations and goals:
% 0.22/0.52
% 0.22/0.52 Axiom 1 (m__901): sdtpldt0(w0_2, w1_2) = xx.
% 0.22/0.52 Axiom 2 (m__901_5): aElementOf0(w1_2, xJ) = true2.
% 0.22/0.52 Axiom 3 (m__901_6): aElementOf0(w0_2, xI) = true2.
% 0.22/0.52
% 0.22/0.52 Goal 1 (m__): tuple(xx, aElementOf0(X, xI), aElementOf0(Y, xJ)) = tuple(sdtpldt0(X, Y), true2, true2).
% 0.22/0.52 The goal is true when:
% 0.22/0.52 X = w0_2
% 0.22/0.52 Y = w1_2
% 0.22/0.52
% 0.22/0.52 Proof:
% 0.22/0.52 tuple(xx, aElementOf0(w0_2, xI), aElementOf0(w1_2, xJ))
% 0.22/0.52 = { by axiom 3 (m__901_6) }
% 0.22/0.52 tuple(xx, true2, aElementOf0(w1_2, xJ))
% 0.22/0.52 = { by axiom 2 (m__901_5) }
% 0.22/0.52 tuple(xx, true2, true2)
% 0.22/0.52 = { by axiom 1 (m__901) R->L }
% 0.22/0.52 tuple(sdtpldt0(w0_2, w1_2), true2, true2)
% 0.22/0.52 % SZS output end Proof
% 0.22/0.52
% 0.22/0.52 RESULT: Theorem (the conjecture is true).
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