TSTP Solution File: NUM854+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM854+1 : TPTP v8.1.2. Released v4.1.0.
% 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 : n026.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 11:58:39 EDT 2023
% Result : Theorem 0.20s 0.49s
% Output : Proof 0.20s
% 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 : NUM854+1 : TPTP v8.1.2. Released v4.1.0.
% 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 : n026.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 15:11:32 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.49 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.49
% 0.20/0.49 % SZS status Theorem
% 0.20/0.49
% 0.20/0.49 % SZS output start Proof
% 0.20/0.49 Take the following subset of the input axioms:
% 0.20/0.49 fof('ass(cond(299, 0), 2)', axiom, ![Vd456, Vd457, Vd458]: (greater(Vd457, Vd458) => greater(vmul(Vd457, Vd456), vmul(Vd458, Vd456)))).
% 0.20/0.49 fof('holds(conjunct1(314), 510, 0)', axiom, greater(vd508, vd509)).
% 0.20/0.49 fof('holds(conjunct1(315), 514, 0)', conjecture, greater(vmul(vd508, vd511), vmul(vd509, vd511))).
% 0.20/0.49
% 0.20/0.49 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.49 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.49 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.49 fresh(y, y, x1...xn) = u
% 0.20/0.50 C => fresh(s, t, x1...xn) = v
% 0.20/0.50 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.50 variables of u and v.
% 0.20/0.50 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.50 input problem has no model of domain size 1).
% 0.20/0.50
% 0.20/0.50 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.50
% 0.20/0.50 Axiom 1 (holds(conjunct1(314), 510, 0)): greater(vd508, vd509) = true2.
% 0.20/0.50 Axiom 2 (ass(cond(299, 0), 2)): fresh26(X, X, Y, Z, W) = true2.
% 0.20/0.50 Axiom 3 (ass(cond(299, 0), 2)): fresh26(greater(X, Y), true2, Z, X, Y) = greater(vmul(X, Z), vmul(Y, Z)).
% 0.20/0.50
% 0.20/0.50 Goal 1 (holds(conjunct1(315), 514, 0)): greater(vmul(vd508, vd511), vmul(vd509, vd511)) = true2.
% 0.20/0.50 Proof:
% 0.20/0.50 greater(vmul(vd508, vd511), vmul(vd509, vd511))
% 0.20/0.50 = { by axiom 3 (ass(cond(299, 0), 2)) R->L }
% 0.20/0.50 fresh26(greater(vd508, vd509), true2, vd511, vd508, vd509)
% 0.20/0.50 = { by axiom 1 (holds(conjunct1(314), 510, 0)) }
% 0.20/0.50 fresh26(true2, true2, vd511, vd508, vd509)
% 0.20/0.50 = { by axiom 2 (ass(cond(299, 0), 2)) }
% 0.20/0.50 true2
% 0.20/0.50 % SZS output end Proof
% 0.20/0.50
% 0.20/0.50 RESULT: Theorem (the conjecture is true).
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