TSTP Solution File: NUM427+3 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM427+3 : 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 : n018.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 11:56:13 EDT 2023
% Result : Theorem 0.20s 0.43s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM427+3 : 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 : n018.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 : Fri Aug 25 16:07:43 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.43 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.20/0.43
% 0.20/0.43 % SZS status Theorem
% 0.20/0.43
% 0.20/0.43 % SZS output start Proof
% 0.20/0.43 Take the following subset of the input axioms:
% 0.20/0.43 fof(mDivisor, definition, ![W0]: (aInteger0(W0) => ![W1]: (aDivisorOf0(W1, W0) <=> (aInteger0(W1) & (W1!=sz00 & ?[W2]: (aInteger0(W2) & sdtasdt0(W1, W2)=W0)))))).
% 0.20/0.43 fof(mIntNeg, axiom, ![W0_2]: (aInteger0(W0_2) => aInteger0(smndt0(W0_2)))).
% 0.20/0.43 fof(m__, conjecture, ?[W0_2]: (aInteger0(W0_2) & sdtasdt0(xq, W0_2)=sdtpldt0(xb, smndt0(xa))) | (aDivisorOf0(xq, sdtpldt0(xb, smndt0(xa))) | sdteqdtlpzmzozddtrp0(xb, xa, xq))).
% 0.20/0.43 fof(m__747, hypothesis, aInteger0(xn) & sdtasdt0(xq, xn)=sdtpldt0(xa, smndt0(xb))).
% 0.20/0.43 fof(m__767, hypothesis, sdtasdt0(xq, smndt0(xn))=sdtpldt0(xb, smndt0(xa))).
% 0.20/0.43
% 0.20/0.43 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.43 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.43 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.43 fresh(y, y, x1...xn) = u
% 0.20/0.43 C => fresh(s, t, x1...xn) = v
% 0.20/0.43 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.43 variables of u and v.
% 0.20/0.43 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.43 input problem has no model of domain size 1).
% 0.20/0.43
% 0.20/0.43 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.43
% 0.20/0.43 Axiom 1 (m__747_1): aInteger0(xn) = true2.
% 0.20/0.43 Axiom 2 (m__767): sdtasdt0(xq, smndt0(xn)) = sdtpldt0(xb, smndt0(xa)).
% 0.20/0.43 Axiom 3 (mIntNeg): fresh15(X, X, Y) = true2.
% 0.20/0.43 Axiom 4 (mIntNeg): fresh15(aInteger0(X), true2, X) = aInteger0(smndt0(X)).
% 0.20/0.43
% 0.20/0.43 Goal 1 (m__): tuple2(sdtasdt0(xq, X), aInteger0(X)) = tuple2(sdtpldt0(xb, smndt0(xa)), true2).
% 0.20/0.43 The goal is true when:
% 0.20/0.43 X = smndt0(xn)
% 0.20/0.43
% 0.20/0.43 Proof:
% 0.20/0.43 tuple2(sdtasdt0(xq, smndt0(xn)), aInteger0(smndt0(xn)))
% 0.20/0.43 = { by axiom 2 (m__767) }
% 0.20/0.43 tuple2(sdtpldt0(xb, smndt0(xa)), aInteger0(smndt0(xn)))
% 0.20/0.43 = { by axiom 4 (mIntNeg) R->L }
% 0.20/0.43 tuple2(sdtpldt0(xb, smndt0(xa)), fresh15(aInteger0(xn), true2, xn))
% 0.20/0.43 = { by axiom 1 (m__747_1) }
% 0.20/0.43 tuple2(sdtpldt0(xb, smndt0(xa)), fresh15(true2, true2, xn))
% 0.20/0.43 = { by axiom 3 (mIntNeg) }
% 0.20/0.43 tuple2(sdtpldt0(xb, smndt0(xa)), true2)
% 0.20/0.43 % SZS output end Proof
% 0.20/0.43
% 0.20/0.43 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------