TSTP Solution File: SYN187-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN187-1 : 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 : n026.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 03:33:31 EDT 2023
% Result : Unsatisfiable 13.37s 2.10s
% Output : Proof 13.37s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.14 % Problem : SYN187-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.15 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.37 % Computer : n026.cluster.edu
% 0.15/0.37 % Model : x86_64 x86_64
% 0.15/0.37 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37 % Memory : 8042.1875MB
% 0.15/0.37 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37 % CPULimit : 300
% 0.15/0.37 % WCLimit : 300
% 0.15/0.37 % DateTime : Sat Aug 26 17:05:48 EDT 2023
% 0.15/0.38 % CPUTime :
% 13.37/2.10 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 13.37/2.10
% 13.37/2.10 % SZS status Unsatisfiable
% 13.37/2.10
% 13.37/2.10 % SZS output start Proof
% 13.37/2.10 Take the following subset of the input axioms:
% 13.37/2.10 fof(axiom_1, axiom, s0(d)).
% 13.37/2.10 fof(axiom_17, axiom, ![X]: q0(X, d)).
% 13.37/2.10 fof(axiom_28, axiom, k0(e)).
% 13.37/2.10 fof(prove_this, negated_conjecture, ~r1(d)).
% 13.37/2.10 fof(rule_117, axiom, q1(d, d, d) | (~k0(e) | ~s0(d))).
% 13.37/2.10 fof(rule_124, axiom, ![D, E]: (r1(D) | (~q0(D, E) | (~s0(d) | ~q1(d, E, d))))).
% 13.37/2.10
% 13.37/2.10 Now clausify the problem and encode Horn clauses using encoding 3 of
% 13.37/2.10 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 13.37/2.10 We repeatedly replace C & s=t => u=v by the two clauses:
% 13.37/2.10 fresh(y, y, x1...xn) = u
% 13.37/2.10 C => fresh(s, t, x1...xn) = v
% 13.37/2.10 where fresh is a fresh function symbol and x1..xn are the free
% 13.37/2.10 variables of u and v.
% 13.37/2.10 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 13.37/2.10 input problem has no model of domain size 1).
% 13.37/2.10
% 13.37/2.10 The encoding turns the above axioms into the following unit equations and goals:
% 13.37/2.10
% 13.37/2.10 Axiom 1 (axiom_1): s0(d) = true.
% 13.37/2.10 Axiom 2 (axiom_28): k0(e) = true.
% 13.37/2.10 Axiom 3 (axiom_17): q0(X, d) = true.
% 13.37/2.10 Axiom 4 (rule_117): fresh285(X, X) = true.
% 13.37/2.10 Axiom 5 (rule_117): fresh286(X, X) = q1(d, d, d).
% 13.37/2.10 Axiom 6 (rule_124): fresh593(X, X, Y) = true.
% 13.37/2.10 Axiom 7 (rule_117): fresh286(k0(e), true) = fresh285(s0(d), true).
% 13.37/2.10 Axiom 8 (rule_124): fresh276(X, X, Y) = r1(Y).
% 13.37/2.10 Axiom 9 (rule_124): fresh592(X, X, Y, Z) = fresh593(s0(d), true, Y).
% 13.37/2.10 Axiom 10 (rule_124): fresh592(q1(d, X, d), true, Y, X) = fresh276(q0(Y, X), true, Y).
% 13.37/2.10
% 13.37/2.10 Goal 1 (prove_this): r1(d) = true.
% 13.37/2.10 Proof:
% 13.37/2.10 r1(d)
% 13.37/2.10 = { by axiom 8 (rule_124) R->L }
% 13.37/2.10 fresh276(true, true, d)
% 13.37/2.10 = { by axiom 3 (axiom_17) R->L }
% 13.37/2.10 fresh276(q0(d, d), true, d)
% 13.37/2.10 = { by axiom 10 (rule_124) R->L }
% 13.37/2.10 fresh592(q1(d, d, d), true, d, d)
% 13.37/2.10 = { by axiom 5 (rule_117) R->L }
% 13.37/2.10 fresh592(fresh286(true, true), true, d, d)
% 13.37/2.10 = { by axiom 2 (axiom_28) R->L }
% 13.37/2.10 fresh592(fresh286(k0(e), true), true, d, d)
% 13.37/2.10 = { by axiom 7 (rule_117) }
% 13.37/2.10 fresh592(fresh285(s0(d), true), true, d, d)
% 13.37/2.10 = { by axiom 1 (axiom_1) }
% 13.37/2.10 fresh592(fresh285(true, true), true, d, d)
% 13.37/2.10 = { by axiom 4 (rule_117) }
% 13.37/2.10 fresh592(true, true, d, d)
% 13.37/2.10 = { by axiom 9 (rule_124) }
% 13.37/2.10 fresh593(s0(d), true, d)
% 13.37/2.10 = { by axiom 1 (axiom_1) }
% 13.37/2.10 fresh593(true, true, d)
% 13.37/2.10 = { by axiom 6 (rule_124) }
% 13.37/2.10 true
% 13.37/2.10 % SZS output end Proof
% 13.37/2.10
% 13.37/2.10 RESULT: Unsatisfiable (the axioms are contradictory).
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