TSTP Solution File: RNG115+4 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : RNG115+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 : n025.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 0.19s 0.70s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : RNG115+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
% 0.13/0.34 % Computer : n025.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 01:41:08 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.70 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.70
% 0.19/0.70 % SZS status Theorem
% 0.19/0.70
% 0.19/0.70 % SZS output start Proof
% 0.19/0.70 Take the following subset of the input axioms:
% 0.19/0.71 fof(m__, conjecture, ?[W0, W1]: ((?[W2_2]: (aElement0(W2_2) & sdtasdt0(xa, W2_2)=W0) | aElementOf0(W0, slsdtgt0(xa))) & ((?[W2]: (aElement0(W2) & sdtasdt0(xb, W2)=W1) | aElementOf0(W1, slsdtgt0(xb))) & xu=sdtpldt0(W0, W1)))).
% 0.19/0.71 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)))))))))).
% 0.19/0.71 fof(m__2273, hypothesis, ?[W0_2, W1_2]: (aElementOf0(W0_2, slsdtgt0(xa)) & (aElementOf0(W1_2, slsdtgt0(xb)) & sdtpldt0(W0_2, W1_2)=xu)) & (aElementOf0(xu, xI) & (xu!=sz00 & ![W0_2]: (((?[W1_2, W2_3]: (aElementOf0(W1_2, slsdtgt0(xa)) & (aElementOf0(W2_3, slsdtgt0(xb)) & sdtpldt0(W1_2, W2_3)=W0_2)) | aElementOf0(W0_2, xI)) & W0_2!=sz00) => ~iLess0(sbrdtbr0(W0_2), sbrdtbr0(xu)))))).
% 0.19/0.71 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))))).
% 0.19/0.71
% 0.19/0.71 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.71 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.71 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.71 fresh(y, y, x1...xn) = u
% 0.19/0.71 C => fresh(s, t, x1...xn) = v
% 0.19/0.71 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.71 variables of u and v.
% 0.19/0.71 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.71 input problem has no model of domain size 1).
% 0.19/0.71
% 0.19/0.71 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.71
% 0.19/0.71 Axiom 1 (m__2273): sdtpldt0(w0, w1) = xu.
% 0.19/0.71 Axiom 2 (m__2273_2): aElementOf0(w1, slsdtgt0(xb)) = true2.
% 0.19/0.71 Axiom 3 (m__2273_3): aElementOf0(w0, slsdtgt0(xa)) = true2.
% 0.19/0.71
% 0.19/0.71 Goal 1 (m___3): tuple4(xu, aElementOf0(X, slsdtgt0(xa)), aElementOf0(Y, slsdtgt0(xb))) = tuple4(sdtpldt0(X, Y), true2, true2).
% 0.19/0.71 The goal is true when:
% 0.19/0.71 X = w0
% 0.19/0.71 Y = w1
% 0.19/0.71
% 0.19/0.71 Proof:
% 0.19/0.71 tuple4(xu, aElementOf0(w0, slsdtgt0(xa)), aElementOf0(w1, slsdtgt0(xb)))
% 0.19/0.71 = { by axiom 3 (m__2273_3) }
% 0.19/0.71 tuple4(xu, true2, aElementOf0(w1, slsdtgt0(xb)))
% 0.19/0.71 = { by axiom 2 (m__2273_2) }
% 0.19/0.71 tuple4(xu, true2, true2)
% 0.19/0.71 = { by axiom 1 (m__2273) R->L }
% 0.19/0.71 tuple4(sdtpldt0(w0, w1), true2, true2)
% 0.19/0.71 % SZS output end Proof
% 0.19/0.71
% 0.19/0.71 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------