TSTP Solution File: NUM490+3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM490+3 : TPTP v8.1.2. Released v4.0.0.
% 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 : n024.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 11:56:35 EDT 2023
% Result : Theorem 0.19s 0.54s
% 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
% 0.06/0.12 % Problem : NUM490+3 : TPTP v8.1.2. Released v4.0.0.
% 0.06/0.13 % 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 : n024.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 17:01:24 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.54 Command-line arguments: --no-flatten-goal
% 0.19/0.54
% 0.19/0.54 % SZS status Theorem
% 0.19/0.54
% 0.19/0.54 % SZS output start Proof
% 0.19/0.54 Take the following subset of the input axioms:
% 0.19/0.54 fof(m__, conjecture, sdtpldt0(sdtasdt0(xp, xm), sdtasdt0(xr, xm))=sdtasdt0(xn, xm) | sdtasdt0(xr, xm)=sdtmndt0(sdtasdt0(xn, xm), sdtasdt0(xp, xm))).
% 0.19/0.54 fof(m__1951, hypothesis, sdtasdt0(xn, xm)=sdtpldt0(sdtasdt0(xp, xm), sdtasdt0(xr, xm))).
% 0.19/0.54
% 0.19/0.54 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.54 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.54 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.54 fresh(y, y, x1...xn) = u
% 0.19/0.54 C => fresh(s, t, x1...xn) = v
% 0.19/0.54 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.54 variables of u and v.
% 0.19/0.54 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.54 input problem has no model of domain size 1).
% 0.19/0.54
% 0.19/0.54 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.54
% 0.19/0.54 Axiom 1 (m__1951): sdtasdt0(xn, xm) = sdtpldt0(sdtasdt0(xp, xm), sdtasdt0(xr, xm)).
% 0.19/0.54
% 0.19/0.54 Goal 1 (m__): sdtpldt0(sdtasdt0(xp, xm), sdtasdt0(xr, xm)) = sdtasdt0(xn, xm).
% 0.19/0.54 Proof:
% 0.19/0.55 sdtpldt0(sdtasdt0(xp, xm), sdtasdt0(xr, xm))
% 0.19/0.55 = { by axiom 1 (m__1951) R->L }
% 0.19/0.55 sdtasdt0(xn, xm)
% 0.19/0.55 % SZS output end Proof
% 0.19/0.55
% 0.19/0.55 RESULT: Theorem (the conjecture is true).
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