TSTP Solution File: SYN238-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN238-1 : 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 : n021.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 03:33:43 EDT 2023
% Result : Unsatisfiable 12.30s 1.96s
% Output : Proof 12.30s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SYN238-1 : 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
% 0.13/0.35 % Computer : n021.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 21:26:12 EDT 2023
% 0.13/0.35 % CPUTime :
% 12.30/1.96 Command-line arguments: --ground-connectedness --complete-subsets
% 12.30/1.96
% 12.30/1.96 % SZS status Unsatisfiable
% 12.30/1.96
% 12.30/1.96 % SZS output start Proof
% 12.30/1.96 Take the following subset of the input axioms:
% 12.30/1.96 fof(axiom_31, axiom, m0(b, b, e)).
% 12.30/1.96 fof(prove_this, negated_conjecture, ![X, Y]: ~m1(X, b, Y)).
% 12.30/1.96 fof(rule_005, axiom, ![C, B]: (m1(B, C, B) | ~m0(C, C, B))).
% 12.30/1.96
% 12.30/1.96 Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.30/1.96 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.30/1.96 We repeatedly replace C & s=t => u=v by the two clauses:
% 12.30/1.96 fresh(y, y, x1...xn) = u
% 12.30/1.96 C => fresh(s, t, x1...xn) = v
% 12.30/1.96 where fresh is a fresh function symbol and x1..xn are the free
% 12.30/1.96 variables of u and v.
% 12.30/1.96 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.30/1.96 input problem has no model of domain size 1).
% 12.30/1.96
% 12.30/1.96 The encoding turns the above axioms into the following unit equations and goals:
% 12.30/1.96
% 12.30/1.96 Axiom 1 (axiom_31): m0(b, b, e) = true2.
% 12.30/1.96 Axiom 2 (rule_005): fresh437(X, X, Y, Z) = true2.
% 12.30/1.96 Axiom 3 (rule_005): fresh437(m0(X, X, Y), true2, Y, X) = m1(Y, X, Y).
% 12.30/1.96
% 12.30/1.96 Goal 1 (prove_this): m1(X, b, Y) = true2.
% 12.30/1.96 The goal is true when:
% 12.30/1.96 X = e
% 12.30/1.96 Y = e
% 12.30/1.96
% 12.30/1.96 Proof:
% 12.30/1.96 m1(e, b, e)
% 12.30/1.96 = { by axiom 3 (rule_005) R->L }
% 12.30/1.96 fresh437(m0(b, b, e), true2, e, b)
% 12.30/1.96 = { by axiom 1 (axiom_31) }
% 12.30/1.96 fresh437(true2, true2, e, b)
% 12.30/1.96 = { by axiom 2 (rule_005) }
% 12.30/1.96 true2
% 12.30/1.96 % SZS output end Proof
% 12.30/1.96
% 12.30/1.96 RESULT: Unsatisfiable (the axioms are contradictory).
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