TSTP Solution File: SYN062-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN062-1 : TPTP v8.1.2. Released v1.0.0.
% 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 : n012.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 03:32:57 EDT 2023
% Result : Unsatisfiable 0.19s 0.38s
% Output : Proof 0.19s
% 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 : SYN062-1 : TPTP v8.1.2. Released v1.0.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n012.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Sat Aug 26 20:27:41 EDT 2023
% 0.19/0.34 % CPUTime :
% 0.19/0.38 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.38
% 0.19/0.38 % SZS status Unsatisfiable
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% 0.19/0.39 % SZS output start Proof
% 0.19/0.39 Take the following subset of the input axioms:
% 0.19/0.39 fof(clause_2, axiom, ![X]: (~big_f(X) | (~big_h(X) | big_i(X)))).
% 0.19/0.39 fof(clause_3, axiom, ![X2]: (~big_i(X2) | (~big_h(X2) | big_j(X2)))).
% 0.19/0.39 fof(clause_4, axiom, ![X2]: (~big_k(X2) | big_h(X2))).
% 0.19/0.39 fof(clause_5, negated_conjecture, big_f(a)).
% 0.19/0.39 fof(clause_6, negated_conjecture, big_k(a)).
% 0.19/0.39 fof(clause_7, negated_conjecture, ~big_j(a)).
% 0.19/0.39
% 0.19/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.39 fresh(y, y, x1...xn) = u
% 0.19/0.39 C => fresh(s, t, x1...xn) = v
% 0.19/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.39 variables of u and v.
% 0.19/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.39 input problem has no model of domain size 1).
% 0.19/0.39
% 0.19/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.39
% 0.19/0.39 Axiom 1 (clause_5): big_f(a) = true.
% 0.19/0.39 Axiom 2 (clause_6): big_k(a) = true.
% 0.19/0.39 Axiom 3 (clause_4): fresh(X, X, Y) = true.
% 0.19/0.39 Axiom 4 (clause_2): fresh5(X, X, Y) = big_i(Y).
% 0.19/0.39 Axiom 5 (clause_2): fresh4(X, X, Y) = true.
% 0.19/0.39 Axiom 6 (clause_3): fresh3(X, X, Y) = big_j(Y).
% 0.19/0.39 Axiom 7 (clause_3): fresh2(X, X, Y) = true.
% 0.19/0.39 Axiom 8 (clause_4): fresh(big_k(X), true, X) = big_h(X).
% 0.19/0.39 Axiom 9 (clause_2): fresh5(big_h(X), true, X) = fresh4(big_f(X), true, X).
% 0.19/0.39 Axiom 10 (clause_3): fresh3(big_h(X), true, X) = fresh2(big_i(X), true, X).
% 0.19/0.39
% 0.19/0.39 Lemma 11: big_h(a) = true.
% 0.19/0.39 Proof:
% 0.19/0.39 big_h(a)
% 0.19/0.39 = { by axiom 8 (clause_4) R->L }
% 0.19/0.39 fresh(big_k(a), true, a)
% 0.19/0.39 = { by axiom 2 (clause_6) }
% 0.19/0.39 fresh(true, true, a)
% 0.19/0.39 = { by axiom 3 (clause_4) }
% 0.19/0.39 true
% 0.19/0.39
% 0.19/0.39 Goal 1 (clause_7): big_j(a) = true.
% 0.19/0.39 Proof:
% 0.19/0.39 big_j(a)
% 0.19/0.39 = { by axiom 6 (clause_3) R->L }
% 0.19/0.39 fresh3(true, true, a)
% 0.19/0.39 = { by lemma 11 R->L }
% 0.19/0.39 fresh3(big_h(a), true, a)
% 0.19/0.39 = { by axiom 10 (clause_3) }
% 0.19/0.39 fresh2(big_i(a), true, a)
% 0.19/0.39 = { by axiom 4 (clause_2) R->L }
% 0.19/0.39 fresh2(fresh5(true, true, a), true, a)
% 0.19/0.39 = { by lemma 11 R->L }
% 0.19/0.39 fresh2(fresh5(big_h(a), true, a), true, a)
% 0.19/0.39 = { by axiom 9 (clause_2) }
% 0.19/0.39 fresh2(fresh4(big_f(a), true, a), true, a)
% 0.19/0.39 = { by axiom 1 (clause_5) }
% 0.19/0.39 fresh2(fresh4(true, true, a), true, a)
% 0.19/0.39 = { by axiom 5 (clause_2) }
% 0.19/0.39 fresh2(true, true, a)
% 0.19/0.39 = { by axiom 7 (clause_3) }
% 0.19/0.39 true
% 0.19/0.39 % SZS output end Proof
% 0.19/0.39
% 0.19/0.39 RESULT: Unsatisfiable (the axioms are contradictory).
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