TSTP Solution File: NUM838+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM838+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 : n021.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 11:58:32 EDT 2023
% Result : Theorem 0.17s 0.52s
% Output : Proof 0.17s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.20 % Problem : NUM838+1 : TPTP v8.1.2. Released v4.1.0.
% 0.07/0.21 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.41 % Computer : n021.cluster.edu
% 0.12/0.41 % Model : x86_64 x86_64
% 0.12/0.41 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.41 % Memory : 8042.1875MB
% 0.12/0.41 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.41 % CPULimit : 300
% 0.12/0.41 % WCLimit : 300
% 0.12/0.41 % DateTime : Fri Aug 25 15:26:56 EDT 2023
% 0.12/0.41 % CPUTime :
% 0.17/0.52 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.17/0.52
% 0.17/0.52 % SZS status Theorem
% 0.17/0.52
% 0.17/0.52 % SZS output start Proof
% 0.17/0.52 Take the following subset of the input axioms:
% 0.17/0.52 fof('ass(cond(proof(196), 0), 4)', axiom, ![Vd310, Vd311, Vd312]: (greater(Vd311, Vd312) => greater(vplus(Vd311, Vd310), vplus(Vd312, Vd310)))).
% 0.17/0.52 fof('holds(antec(195), 304, 0)', axiom, greater(vd301, vd302)).
% 0.17/0.52 fof('holds(conseq(195), 305, 0)', conjecture, greater(vplus(vd301, vd303), vplus(vd302, vd303))).
% 0.17/0.52
% 0.17/0.52 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.17/0.52 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.17/0.52 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.17/0.52 fresh(y, y, x1...xn) = u
% 0.17/0.52 C => fresh(s, t, x1...xn) = v
% 0.17/0.52 where fresh is a fresh function symbol and x1..xn are the free
% 0.17/0.52 variables of u and v.
% 0.17/0.52 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.17/0.52 input problem has no model of domain size 1).
% 0.17/0.52
% 0.17/0.52 The encoding turns the above axioms into the following unit equations and goals:
% 0.17/0.52
% 0.17/0.52 Axiom 1 (holds(antec(195), 304, 0)): greater(vd301, vd302) = true2.
% 0.17/0.52 Axiom 2 (ass(cond(proof(196), 0), 4)): fresh15(X, X, Y, Z, W) = true2.
% 0.17/0.52 Axiom 3 (ass(cond(proof(196), 0), 4)): fresh15(greater(X, Y), true2, Z, X, Y) = greater(vplus(X, Z), vplus(Y, Z)).
% 0.17/0.52
% 0.17/0.52 Goal 1 (holds(conseq(195), 305, 0)): greater(vplus(vd301, vd303), vplus(vd302, vd303)) = true2.
% 0.17/0.52 Proof:
% 0.17/0.52 greater(vplus(vd301, vd303), vplus(vd302, vd303))
% 0.17/0.52 = { by axiom 3 (ass(cond(proof(196), 0), 4)) R->L }
% 0.17/0.52 fresh15(greater(vd301, vd302), true2, vd303, vd301, vd302)
% 0.17/0.52 = { by axiom 1 (holds(antec(195), 304, 0)) }
% 0.17/0.52 fresh15(true2, true2, vd303, vd301, vd302)
% 0.17/0.52 = { by axiom 2 (ass(cond(proof(196), 0), 4)) }
% 0.17/0.52 true2
% 0.17/0.52 % SZS output end Proof
% 0.17/0.52
% 0.17/0.52 RESULT: Theorem (the conjecture is true).
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