TSTP Solution File: NUM850+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM850+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 : n004.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 11:58:38 EDT 2023
% Result : Theorem 0.18s 0.45s
% Output : Proof 0.18s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11 % Problem : NUM850+1 : TPTP v8.1.2. Released v4.1.0.
% 0.11/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33 % Computer : n004.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Fri Aug 25 09:12:08 EDT 2023
% 0.12/0.33 % CPUTime :
% 0.18/0.45 Command-line arguments: --no-flatten-goal
% 0.18/0.45
% 0.18/0.45 % SZS status Theorem
% 0.18/0.45
% 0.18/0.45 % SZS output start Proof
% 0.18/0.45 Take the following subset of the input axioms:
% 0.18/0.45 fof('ass(cond(12, 0), 0)', axiom, ![Vd16]: vsucc(Vd16)!=Vd16).
% 0.18/0.45 fof('ass(cond(73, 0), 0)', axiom, ![Vd92, Vd93]: Vd93!=vplus(Vd92, Vd93)).
% 0.18/0.45 fof('ass(cond(goal(130), 0), 1)', axiom, ![Vd203, Vd204]: (Vd203!=Vd204 | ~less(Vd203, Vd204))).
% 0.18/0.45 fof('ass(cond(goal(130), 0), 2)', axiom, ![Vd203_2, Vd204_2]: (~greater(Vd203_2, Vd204_2) | ~less(Vd203_2, Vd204_2))).
% 0.18/0.45 fof('ass(cond(goal(130), 0), 3)', axiom, ![Vd203_2, Vd204_2]: (Vd203_2!=Vd204_2 | ~greater(Vd203_2, Vd204_2))).
% 0.18/0.45 fof('ass(cond(goal(88), 0), 1)', axiom, ![Vd120, Vd121]: (Vd120!=Vd121 | ~?[Vd125]: Vd121=vplus(Vd120, Vd125))).
% 0.18/0.45 fof('ass(cond(goal(88), 0), 2)', axiom, ![Vd120_2, Vd121_2]: (~?[Vd123]: Vd120_2=vplus(Vd121_2, Vd123) | ~?[Vd125_2]: Vd121_2=vplus(Vd120_2, Vd125_2))).
% 0.18/0.45 fof('ass(cond(goal(88), 0), 3)', axiom, ![Vd120_2, Vd121_2]: (Vd120_2!=Vd121_2 | ~?[Vd123_2]: Vd120_2=vplus(Vd121_2, Vd123_2))).
% 0.18/0.45 fof('def(cond(conseq(axiom(3)), 11), 1)', axiom, ![Vd193, Vd194]: (greater(Vd194, Vd193) <=> ?[Vd196]: Vd194=vplus(Vd193, Vd196))).
% 0.18/0.45 fof('holds(antec(302), 472, 0)', axiom, greater(vd470, vd471)).
% 0.18/0.45 fof('qe(conseq_conjunct1(conseq(302)))', conjecture, ?[Vd473]: vd470=vplus(vd471, Vd473)).
% 0.18/0.45 fof('qu(restrictor(axiom(1)), holds(scope(axiom(1)), 2, 0))', axiom, ![Vd1]: vsucc(Vd1)!=v1).
% 0.18/0.45
% 0.18/0.45 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.18/0.45 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.18/0.45 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.18/0.45 fresh(y, y, x1...xn) = u
% 0.18/0.45 C => fresh(s, t, x1...xn) = v
% 0.18/0.45 where fresh is a fresh function symbol and x1..xn are the free
% 0.18/0.45 variables of u and v.
% 0.18/0.45 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.18/0.45 input problem has no model of domain size 1).
% 0.18/0.45
% 0.18/0.45 The encoding turns the above axioms into the following unit equations and goals:
% 0.18/0.45
% 0.18/0.45 Axiom 1 (holds(antec(302), 472, 0)): greater(vd470, vd471) = true2.
% 0.18/0.45 Axiom 2 (def(cond(conseq(axiom(3)), 11), 1)_1): fresh5(X, X, Y, Z) = Z.
% 0.18/0.45 Axiom 3 (def(cond(conseq(axiom(3)), 11), 1)_1): fresh5(greater(X, Y), true2, Y, X) = vplus(Y, vd196(Y, X)).
% 0.18/0.45
% 0.18/0.45 Goal 1 (qe(conseq_conjunct1(conseq(302)))): vd470 = vplus(vd471, X).
% 0.18/0.45 The goal is true when:
% 0.18/0.45 X = vd196(vd471, vd470)
% 0.18/0.45
% 0.18/0.45 Proof:
% 0.18/0.45 vd470
% 0.18/0.45 = { by axiom 2 (def(cond(conseq(axiom(3)), 11), 1)_1) R->L }
% 0.18/0.45 fresh5(true2, true2, vd471, vd470)
% 0.18/0.45 = { by axiom 1 (holds(antec(302), 472, 0)) R->L }
% 0.18/0.45 fresh5(greater(vd470, vd471), true2, vd471, vd470)
% 0.18/0.45 = { by axiom 3 (def(cond(conseq(axiom(3)), 11), 1)_1) }
% 0.18/0.45 vplus(vd471, vd196(vd471, vd470))
% 0.18/0.45 % SZS output end Proof
% 0.18/0.45
% 0.18/0.45 RESULT: Theorem (the conjecture is true).
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