TSTP Solution File: SYN251-1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SYN251-1 : 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 : n005.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 03:33:46 EDT 2023
% Result : Unsatisfiable 21.70s 3.20s
% Output : Proof 21.70s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.14 % Problem : SYN251-1 : TPTP v8.1.2. Released v1.1.0.
% 0.08/0.15 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.16/0.37 % Computer : n005.cluster.edu
% 0.16/0.37 % Model : x86_64 x86_64
% 0.16/0.37 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.16/0.37 % Memory : 8042.1875MB
% 0.16/0.37 % OS : Linux 3.10.0-693.el7.x86_64
% 0.16/0.37 % CPULimit : 300
% 0.16/0.37 % WCLimit : 300
% 0.16/0.37 % DateTime : Sat Aug 26 21:42:53 EDT 2023
% 0.16/0.37 % CPUTime :
% 21.70/3.20 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 21.70/3.20
% 21.70/3.20 % SZS status Unsatisfiable
% 21.70/3.20
% 21.70/3.20 % SZS output start Proof
% 21.70/3.20 Take the following subset of the input axioms:
% 21.70/3.20 fof(axiom_19, axiom, ![X, Y]: m0(X, d, Y)).
% 21.70/3.20 fof(axiom_30, axiom, n0(e, e)).
% 21.70/3.20 fof(axiom_32, axiom, k0(b)).
% 21.70/3.20 fof(prove_this, negated_conjecture, ~m1(e, b, b)).
% 21.70/3.20 fof(rule_005, axiom, ![C, B]: (m1(B, C, B) | ~m0(C, C, B))).
% 21.70/3.20 fof(rule_011, axiom, ![J, A2]: (m1(J, J, A2) | (~k0(J) | ~n0(A2, A2)))).
% 21.70/3.20 fof(rule_016, axiom, ![I, H, A, J2]: (m1(H, I, I) | (~m1(J2, I, H) | ~m1(J2, A, I)))).
% 21.70/3.20
% 21.70/3.20 Now clausify the problem and encode Horn clauses using encoding 3 of
% 21.70/3.20 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 21.70/3.20 We repeatedly replace C & s=t => u=v by the two clauses:
% 21.70/3.20 fresh(y, y, x1...xn) = u
% 21.70/3.20 C => fresh(s, t, x1...xn) = v
% 21.70/3.20 where fresh is a fresh function symbol and x1..xn are the free
% 21.70/3.20 variables of u and v.
% 21.70/3.20 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 21.70/3.20 input problem has no model of domain size 1).
% 21.70/3.20
% 21.70/3.20 The encoding turns the above axioms into the following unit equations and goals:
% 21.70/3.20
% 21.70/3.20 Axiom 1 (axiom_30): n0(e, e) = true.
% 21.70/3.20 Axiom 2 (axiom_32): k0(b) = true.
% 21.70/3.20 Axiom 3 (axiom_19): m0(X, d, Y) = true.
% 21.70/3.20 Axiom 4 (rule_005): fresh437(X, X, Y, Z) = true.
% 21.70/3.20 Axiom 5 (rule_011): fresh430(X, X, Y, Z) = m1(Y, Y, Z).
% 21.70/3.20 Axiom 6 (rule_011): fresh429(X, X, Y, Z) = true.
% 21.70/3.20 Axiom 7 (rule_016): fresh422(X, X, Y, Z) = true.
% 21.70/3.20 Axiom 8 (rule_011): fresh430(k0(X), true, X, Y) = fresh429(n0(Y, Y), true, X, Y).
% 21.70/3.20 Axiom 9 (rule_016): fresh423(X, X, Y, Z, W) = m1(Y, Z, Z).
% 21.70/3.20 Axiom 10 (rule_005): fresh437(m0(X, X, Y), true, Y, X) = m1(Y, X, Y).
% 21.70/3.20 Axiom 11 (rule_016): fresh423(m1(X, Y, Z), true, W, Z, X) = fresh422(m1(X, Z, W), true, W, Z).
% 21.70/3.20
% 21.70/3.20 Goal 1 (prove_this): m1(e, b, b) = true.
% 21.70/3.20 Proof:
% 21.70/3.20 m1(e, b, b)
% 21.70/3.20 = { by axiom 9 (rule_016) R->L }
% 21.70/3.20 fresh423(true, true, e, b, b)
% 21.70/3.20 = { by axiom 4 (rule_005) R->L }
% 21.70/3.20 fresh423(fresh437(true, true, b, d), true, e, b, b)
% 21.70/3.20 = { by axiom 3 (axiom_19) R->L }
% 21.70/3.20 fresh423(fresh437(m0(d, d, b), true, b, d), true, e, b, b)
% 21.70/3.20 = { by axiom 10 (rule_005) }
% 21.70/3.20 fresh423(m1(b, d, b), true, e, b, b)
% 21.70/3.20 = { by axiom 11 (rule_016) }
% 21.70/3.20 fresh422(m1(b, b, e), true, e, b)
% 21.70/3.20 = { by axiom 5 (rule_011) R->L }
% 21.70/3.20 fresh422(fresh430(true, true, b, e), true, e, b)
% 21.70/3.20 = { by axiom 2 (axiom_32) R->L }
% 21.70/3.20 fresh422(fresh430(k0(b), true, b, e), true, e, b)
% 21.70/3.20 = { by axiom 8 (rule_011) }
% 21.70/3.20 fresh422(fresh429(n0(e, e), true, b, e), true, e, b)
% 21.70/3.20 = { by axiom 1 (axiom_30) }
% 21.70/3.20 fresh422(fresh429(true, true, b, e), true, e, b)
% 21.70/3.20 = { by axiom 6 (rule_011) }
% 21.70/3.20 fresh422(true, true, e, b)
% 21.70/3.20 = { by axiom 7 (rule_016) }
% 21.70/3.20 true
% 21.70/3.20 % SZS output end Proof
% 21.70/3.20
% 21.70/3.20 RESULT: Unsatisfiable (the axioms are contradictory).
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