TSTP Solution File: SYN130-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN130-1 : TPTP v8.1.2. Released v1.1.0.
% 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 : n029.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 : Fri Sep 1 03:33:17 EDT 2023
% Result : Unsatisfiable 11.64s 1.88s
% Output : Proof 11.64s
% 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 : SYN130-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.17/0.35 % Computer : n029.cluster.edu
% 0.17/0.35 % Model : x86_64 x86_64
% 0.17/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.17/0.35 % Memory : 8042.1875MB
% 0.17/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.17/0.35 % CPULimit : 300
% 0.17/0.35 % WCLimit : 300
% 0.17/0.35 % DateTime : Sat Aug 26 21:51:25 EDT 2023
% 0.17/0.35 % CPUTime :
% 11.64/1.88 Command-line arguments: --ground-connectedness --complete-subsets
% 11.64/1.88
% 11.64/1.88 % SZS status Unsatisfiable
% 11.64/1.88
% 11.64/1.88 % SZS output start Proof
% 11.64/1.88 Take the following subset of the input axioms:
% 11.64/1.88 fof(axiom_19, axiom, ![X, Y]: m0(X, d, Y)).
% 11.64/1.88 fof(prove_this, negated_conjecture, ![X2, Y2]: ~m0(X2, Y2, b)).
% 11.64/1.88
% 11.64/1.88 Now clausify the problem and encode Horn clauses using encoding 3 of
% 11.64/1.88 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 11.64/1.88 We repeatedly replace C & s=t => u=v by the two clauses:
% 11.64/1.88 fresh(y, y, x1...xn) = u
% 11.64/1.88 C => fresh(s, t, x1...xn) = v
% 11.64/1.88 where fresh is a fresh function symbol and x1..xn are the free
% 11.64/1.88 variables of u and v.
% 11.64/1.88 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 11.64/1.88 input problem has no model of domain size 1).
% 11.64/1.88
% 11.64/1.88 The encoding turns the above axioms into the following unit equations and goals:
% 11.64/1.88
% 11.64/1.88 Axiom 1 (axiom_19): m0(X, d, Y) = true2.
% 11.64/1.88
% 11.64/1.88 Goal 1 (prove_this): m0(X, Y, b) = true2.
% 11.64/1.88 The goal is true when:
% 11.64/1.88 X = X
% 11.64/1.88 Y = d
% 11.64/1.88
% 11.64/1.88 Proof:
% 11.64/1.88 m0(X, d, b)
% 11.64/1.88 = { by axiom 1 (axiom_19) }
% 11.64/1.88 true2
% 11.64/1.88 % SZS output end Proof
% 11.64/1.88
% 11.64/1.88 RESULT: Unsatisfiable (the axioms are contradictory).
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