TSTP Solution File: SYN281-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN281-1 : 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 : n007.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 03:33:54 EDT 2023
% Result : Unsatisfiable 13.05s 2.02s
% Output : Proof 13.05s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SYN281-1 : TPTP v8.1.2. Released v1.1.0.
% 0.13/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n007.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Sat Aug 26 17:41:56 EDT 2023
% 0.13/0.35 % CPUTime :
% 13.05/2.02 Command-line arguments: --ground-connectedness --complete-subsets
% 13.05/2.02
% 13.05/2.02 % SZS status Unsatisfiable
% 13.05/2.02
% 13.05/2.02 % SZS output start Proof
% 13.05/2.02 Take the following subset of the input axioms:
% 13.05/2.02 fof(axiom_18, axiom, p0(c, b)).
% 13.05/2.02 fof(prove_this, negated_conjecture, ![X]: ~q1(X, c, c)).
% 13.05/2.02 fof(rule_095, axiom, ![G, F]: (q1(F, G, G) | ~p0(G, F))).
% 13.05/2.02
% 13.05/2.02 Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.05/2.02 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.05/2.02 We repeatedly replace C & s=t => u=v by the two clauses:
% 13.05/2.02 fresh(y, y, x1...xn) = u
% 13.05/2.02 C => fresh(s, t, x1...xn) = v
% 13.05/2.02 where fresh is a fresh function symbol and x1..xn are the free
% 13.05/2.02 variables of u and v.
% 13.05/2.02 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.05/2.02 input problem has no model of domain size 1).
% 13.05/2.02
% 13.05/2.02 The encoding turns the above axioms into the following unit equations and goals:
% 13.05/2.02
% 13.05/2.02 Axiom 1 (axiom_18): p0(c, b) = true2.
% 13.05/2.02 Axiom 2 (rule_095): fresh314(X, X, Y, Z) = true2.
% 13.05/2.02 Axiom 3 (rule_095): fresh314(p0(X, Y), true2, Y, X) = q1(Y, X, X).
% 13.05/2.02
% 13.05/2.02 Goal 1 (prove_this): q1(X, c, c) = true2.
% 13.05/2.02 The goal is true when:
% 13.05/2.02 X = b
% 13.05/2.02
% 13.05/2.02 Proof:
% 13.05/2.02 q1(b, c, c)
% 13.05/2.02 = { by axiom 3 (rule_095) R->L }
% 13.05/2.02 fresh314(p0(c, b), true2, b, c)
% 13.05/2.02 = { by axiom 1 (axiom_18) }
% 13.05/2.02 fresh314(true2, true2, b, c)
% 13.05/2.02 = { by axiom 2 (rule_095) }
% 13.05/2.02 true2
% 13.05/2.02 % SZS output end Proof
% 13.05/2.02
% 13.05/2.02 RESULT: Unsatisfiable (the axioms are contradictory).
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