TSTP Solution File: NUM848+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM848+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 : 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 11:58:37 EDT 2023
% Result : Theorem 0.19s 0.47s
% 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 : NUM848+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 : n018.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 15:53:13 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.47 Command-line arguments: --no-flatten-goal
% 0.19/0.47
% 0.19/0.47 % SZS status Theorem
% 0.19/0.47
% 0.19/0.47 % SZS output start Proof
% 0.19/0.47 Take the following subset of the input axioms:
% 0.19/0.47 fof('ass(cond(261, 0), 0)', axiom, ![Vd408, Vd409]: vmul(vsucc(Vd408), Vd409)=vplus(vmul(Vd408, Vd409), Vd409)).
% 0.19/0.47 fof('ass(cond(270, 0), 0)', axiom, ![Vd418, Vd419]: vmul(Vd418, Vd419)=vmul(Vd419, Vd418)).
% 0.19/0.47 fof('holds(286, 441, 4)', conjecture, vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436))=vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439)))).
% 0.19/0.47
% 0.19/0.47 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.47 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.47 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.47 fresh(y, y, x1...xn) = u
% 0.19/0.47 C => fresh(s, t, x1...xn) = v
% 0.19/0.47 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.47 variables of u and v.
% 0.19/0.47 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.47 input problem has no model of domain size 1).
% 0.19/0.47
% 0.19/0.47 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.47
% 0.19/0.47 Axiom 1 (ass(cond(270, 0), 0)): vmul(X, Y) = vmul(Y, X).
% 0.19/0.47 Axiom 2 (ass(cond(261, 0), 0)): vmul(vsucc(X), Y) = vplus(vmul(X, Y), Y).
% 0.19/0.47
% 0.19/0.47 Goal 1 (holds(286, 441, 4)): vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436)) = vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439))).
% 0.19/0.47 Proof:
% 0.19/0.47 vplus(vmul(vd436, vd437), vplus(vmul(vd436, vd439), vd436))
% 0.19/0.47 = { by axiom 1 (ass(cond(270, 0), 0)) }
% 0.19/0.47 vplus(vmul(vd437, vd436), vplus(vmul(vd436, vd439), vd436))
% 0.19/0.47 = { by axiom 1 (ass(cond(270, 0), 0)) }
% 0.19/0.47 vplus(vmul(vd437, vd436), vplus(vmul(vd439, vd436), vd436))
% 0.19/0.47 = { by axiom 2 (ass(cond(261, 0), 0)) R->L }
% 0.19/0.47 vplus(vmul(vd437, vd436), vmul(vsucc(vd439), vd436))
% 0.19/0.47 = { by axiom 1 (ass(cond(270, 0), 0)) }
% 0.19/0.47 vplus(vmul(vd437, vd436), vmul(vd436, vsucc(vd439)))
% 0.19/0.47 = { by axiom 1 (ass(cond(270, 0), 0)) R->L }
% 0.19/0.47 vplus(vmul(vd436, vd437), vmul(vd436, vsucc(vd439)))
% 0.19/0.47 % SZS output end Proof
% 0.19/0.47
% 0.19/0.48 RESULT: Theorem (the conjecture is true).
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