TSTP Solution File: RNG116+4 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : RNG116+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 : n027.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 13:59:23 EDT 2023
% Result : Theorem 63.81s 8.65s
% Output : Proof 63.81s
% 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 : RNG116+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
% 0.12/0.34 % Computer : n027.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Sun Aug 27 02:41:20 EDT 2023
% 0.12/0.34 % CPUTime :
% 63.81/8.65 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 63.81/8.65
% 63.81/8.65 % SZS status Theorem
% 63.81/8.65
% 63.81/8.65 % SZS output start Proof
% 63.81/8.65 Take the following subset of the input axioms:
% 63.81/8.66 fof(m__, conjecture, ?[W0, W1]: (aElement0(W0) & (aElement0(W1) & xu=sdtpldt0(sdtasdt0(xa, W0), sdtasdt0(xb, W1))))).
% 63.81/8.66 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)))))))))).
% 63.81/8.66 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))))).
% 63.81/8.66 fof(m__2456, hypothesis, ?[W0_2]: (aElement0(W0_2) & sdtasdt0(xa, W0_2)=xk) & (aElementOf0(xk, slsdtgt0(xa)) & (?[W0_2]: (aElement0(W0_2) & sdtasdt0(xb, W0_2)=xl) & (aElementOf0(xl, slsdtgt0(xb)) & xu=sdtpldt0(xk, xl))))).
% 63.81/8.66
% 63.81/8.66 Now clausify the problem and encode Horn clauses using encoding 3 of
% 63.81/8.66 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 63.81/8.66 We repeatedly replace C & s=t => u=v by the two clauses:
% 63.81/8.66 fresh(y, y, x1...xn) = u
% 63.81/8.66 C => fresh(s, t, x1...xn) = v
% 63.81/8.66 where fresh is a fresh function symbol and x1..xn are the free
% 63.81/8.66 variables of u and v.
% 63.81/8.66 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 63.81/8.66 input problem has no model of domain size 1).
% 63.81/8.66
% 63.81/8.66 The encoding turns the above axioms into the following unit equations and goals:
% 63.81/8.66
% 63.81/8.66 Axiom 1 (m__2456_3): aElement0(w0_2) = true2.
% 63.81/8.66 Axiom 2 (m__2456_4): aElement0(w0) = true2.
% 63.81/8.66 Axiom 3 (m__2456_2): sdtasdt0(xb, w0) = xl.
% 63.81/8.66 Axiom 4 (m__2456_1): sdtasdt0(xa, w0_2) = xk.
% 63.81/8.66 Axiom 5 (m__2456): xu = sdtpldt0(xk, xl).
% 63.81/8.66
% 63.81/8.66 Goal 1 (m__): tuple4(xu, aElement0(X), aElement0(Y)) = tuple4(sdtpldt0(sdtasdt0(xa, X), sdtasdt0(xb, Y)), true2, true2).
% 63.81/8.66 The goal is true when:
% 63.81/8.66 X = w0_2
% 63.81/8.66 Y = w0
% 63.81/8.66
% 63.81/8.66 Proof:
% 63.81/8.66 tuple4(xu, aElement0(w0_2), aElement0(w0))
% 63.81/8.66 = { by axiom 1 (m__2456_3) }
% 63.81/8.66 tuple4(xu, true2, aElement0(w0))
% 63.81/8.66 = { by axiom 2 (m__2456_4) }
% 63.81/8.66 tuple4(xu, true2, true2)
% 63.81/8.66 = { by axiom 5 (m__2456) }
% 63.81/8.66 tuple4(sdtpldt0(xk, xl), true2, true2)
% 63.81/8.66 = { by axiom 3 (m__2456_2) R->L }
% 63.81/8.66 tuple4(sdtpldt0(xk, sdtasdt0(xb, w0)), true2, true2)
% 63.81/8.66 = { by axiom 4 (m__2456_1) R->L }
% 63.81/8.66 tuple4(sdtpldt0(sdtasdt0(xa, w0_2), sdtasdt0(xb, w0)), true2, true2)
% 63.81/8.66 % SZS output end Proof
% 63.81/8.66
% 63.81/8.66 RESULT: Theorem (the conjecture is true).
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