TSTP Solution File: NUM849+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM849+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 : n019.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 11:58:37 EDT 2023
% Result : Theorem 0.19s 0.50s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM849+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
% 0.13/0.34 % Computer : n019.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 : Fri Aug 25 14:09:58 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.50 Command-line arguments: --flatten
% 0.19/0.50
% 0.19/0.50 % SZS status Theorem
% 0.19/0.50
% 0.19/0.51 % SZS output start Proof
% 0.19/0.51 Take the following subset of the input axioms:
% 0.19/0.51 fof('ass(cond(281, 0), 0)', axiom, ![Vd432, Vd433, Vd434]: vmul(Vd432, vplus(Vd433, Vd434))=vplus(vmul(Vd432, Vd433), vmul(Vd432, Vd434))).
% 0.19/0.51 fof('ass(cond(61, 0), 0)', axiom, ![Vd78, Vd79]: vplus(Vd79, Vd78)=vplus(Vd78, Vd79)).
% 0.19/0.51 fof('ass(cond(conseq(292), 1), 2)', axiom, ![Vd451]: (vmul(vmul(vd448, vd449), Vd451)=vmul(vd448, vmul(vd449, Vd451)) => vplus(vmul(vmul(vd448, vd449), Vd451), vmul(vd448, vd449))=vplus(vmul(vd448, vmul(vd449, Vd451)), vmul(vd448, vd449)))).
% 0.19/0.51 fof('qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))', axiom, ![Vd396, Vd397]: (vmul(Vd396, vsucc(Vd397))=vplus(vmul(Vd396, Vd397), Vd396) & vmul(Vd396, v1)=Vd396)).
% 0.19/0.51 fof('qu(ind(296), imp(296))', conjecture, ![Vd454]: (vmul(vmul(vd448, vd449), Vd454)=vmul(vd448, vmul(vd449, Vd454)) => vmul(vmul(vd448, vd449), vsucc(Vd454))=vmul(vd448, vmul(vd449, vsucc(Vd454))))).
% 0.19/0.51
% 0.19/0.51 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.51 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.51 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.51 fresh(y, y, x1...xn) = u
% 0.19/0.51 C => fresh(s, t, x1...xn) = v
% 0.19/0.51 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.51 variables of u and v.
% 0.19/0.51 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.51 input problem has no model of domain size 1).
% 0.19/0.51
% 0.19/0.51 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.51
% 0.19/0.51 Axiom 1 (ass(cond(61, 0), 0)): vplus(X, Y) = vplus(Y, X).
% 0.19/0.51 Axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))): vmul(X, v1) = X.
% 0.19/0.51 Axiom 3 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))_1): vmul(X, vsucc(Y)) = vplus(vmul(X, Y), X).
% 0.19/0.51 Axiom 4 (qu(ind(296), imp(296))): vmul(vmul(vd448, vd449), vd454) = vmul(vd448, vmul(vd449, vd454)).
% 0.19/0.51 Axiom 5 (ass(cond(281, 0), 0)): vmul(X, vplus(Y, Z)) = vplus(vmul(X, Y), vmul(X, Z)).
% 0.19/0.51 Axiom 6 (ass(cond(conseq(292), 1), 2)): fresh24(X, X, Y) = vplus(vmul(vd448, vmul(vd449, Y)), vmul(vd448, vd449)).
% 0.19/0.51 Axiom 7 (ass(cond(conseq(292), 1), 2)): fresh24(vmul(vmul(vd448, vd449), X), vmul(vd448, vmul(vd449, X)), X) = vplus(vmul(vmul(vd448, vd449), X), vmul(vd448, vd449)).
% 0.19/0.51
% 0.19/0.51 Lemma 8: vplus(X, vmul(X, Y)) = vmul(X, vsucc(Y)).
% 0.19/0.51 Proof:
% 0.19/0.51 vplus(X, vmul(X, Y))
% 0.19/0.51 = { by axiom 1 (ass(cond(61, 0), 0)) R->L }
% 0.19/0.51 vplus(vmul(X, Y), X)
% 0.19/0.51 = { by axiom 3 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))_1) R->L }
% 0.19/0.51 vmul(X, vsucc(Y))
% 0.19/0.51
% 0.19/0.51 Lemma 9: vmul(vmul(X, Y), v1) = vmul(X, vmul(Y, v1)).
% 0.19/0.51 Proof:
% 0.19/0.51 vmul(vmul(X, Y), v1)
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) }
% 0.19/0.51 vmul(X, Y)
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) R->L }
% 0.19/0.51 vmul(X, vmul(Y, v1))
% 0.19/0.51
% 0.19/0.51 Goal 1 (qu(ind(296), imp(296))_1): vmul(vmul(vd448, vd449), vsucc(vd454)) = vmul(vd448, vmul(vd449, vsucc(vd454))).
% 0.19/0.51 Proof:
% 0.19/0.51 vmul(vmul(vd448, vd449), vsucc(vd454))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) R->L }
% 0.19/0.51 vmul(vmul(vd448, vmul(vd449, v1)), vsucc(vd454))
% 0.19/0.51 = { by lemma 9 R->L }
% 0.19/0.51 vmul(vmul(vmul(vd448, vd449), v1), vsucc(vd454))
% 0.19/0.51 = { by lemma 8 R->L }
% 0.19/0.51 vplus(vmul(vmul(vd448, vd449), v1), vmul(vmul(vmul(vd448, vd449), v1), vd454))
% 0.19/0.51 = { by axiom 1 (ass(cond(61, 0), 0)) R->L }
% 0.19/0.51 vplus(vmul(vmul(vmul(vd448, vd449), v1), vd454), vmul(vmul(vd448, vd449), v1))
% 0.19/0.51 = { by lemma 9 }
% 0.19/0.51 vplus(vmul(vmul(vmul(vd448, vd449), v1), vd454), vmul(vd448, vmul(vd449, v1)))
% 0.19/0.51 = { by lemma 9 }
% 0.19/0.51 vplus(vmul(vmul(vd448, vmul(vd449, v1)), vd454), vmul(vd448, vmul(vd449, v1)))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) }
% 0.19/0.51 vplus(vmul(vmul(vd448, vmul(vd449, v1)), vd454), vmul(vd448, vd449))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) }
% 0.19/0.51 vplus(vmul(vmul(vd448, vd449), vd454), vmul(vd448, vd449))
% 0.19/0.51 = { by axiom 7 (ass(cond(conseq(292), 1), 2)) R->L }
% 0.19/0.51 fresh24(vmul(vmul(vd448, vd449), vd454), vmul(vd448, vmul(vd449, vd454)), vd454)
% 0.19/0.51 = { by axiom 4 (qu(ind(296), imp(296))) R->L }
% 0.19/0.51 fresh24(vmul(vmul(vd448, vd449), vd454), vmul(vmul(vd448, vd449), vd454), vd454)
% 0.19/0.51 = { by axiom 6 (ass(cond(conseq(292), 1), 2)) }
% 0.19/0.51 vplus(vmul(vd448, vmul(vd449, vd454)), vmul(vd448, vd449))
% 0.19/0.51 = { by axiom 5 (ass(cond(281, 0), 0)) R->L }
% 0.19/0.51 vmul(vd448, vplus(vmul(vd449, vd454), vd449))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) R->L }
% 0.19/0.51 vmul(vd448, vplus(vmul(vmul(vd449, v1), vd454), vd449))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) R->L }
% 0.19/0.51 vmul(vd448, vplus(vmul(vmul(vd449, v1), vd454), vmul(vd449, v1)))
% 0.19/0.51 = { by axiom 1 (ass(cond(61, 0), 0)) }
% 0.19/0.51 vmul(vd448, vplus(vmul(vd449, v1), vmul(vmul(vd449, v1), vd454)))
% 0.19/0.51 = { by lemma 8 }
% 0.19/0.51 vmul(vd448, vmul(vmul(vd449, v1), vsucc(vd454)))
% 0.19/0.51 = { by axiom 2 (qu(cond(conseq(axiom(3)), 32), and(holds(definiens(249), 399, 0), holds(definiens(249), 398, 0)))) }
% 0.19/0.51 vmul(vd448, vmul(vd449, vsucc(vd454)))
% 0.19/0.51 % SZS output end Proof
% 0.19/0.51
% 0.19/0.51 RESULT: Theorem (the conjecture is true).
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