TSTP Solution File: SWC385-1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC385-1 : TPTP v8.1.2. Released v2.4.0.
% 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 : Thu Aug 31 20:55:18 EDT 2023
% Result : Unsatisfiable 3.48s 0.87s
% Output : Proof 3.48s
% 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 : SWC385-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35 % Computer : n012.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Mon Aug 28 15:02:09 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.48/0.87 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 3.48/0.87
% 3.48/0.87 % SZS status Unsatisfiable
% 3.48/0.87
% 3.48/0.87 % SZS output start Proof
% 3.48/0.87 Take the following subset of the input axioms:
% 3.48/0.87 fof(co1_10, negated_conjecture, ~singletonP(sk1) | ~segmentP(sk2, sk1)).
% 3.48/0.87 fof(co1_5, negated_conjecture, sk2=sk4).
% 3.48/0.87 fof(co1_6, negated_conjecture, sk1=sk3).
% 3.48/0.87 fof(co1_7, negated_conjecture, neq(sk2, nil)).
% 3.48/0.87 fof(co1_8, negated_conjecture, segmentP(sk4, sk3)).
% 3.48/0.87 fof(co1_9, negated_conjecture, singletonP(sk3) | ~neq(sk4, nil)).
% 3.48/0.87
% 3.48/0.87 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.48/0.87 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.48/0.87 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.48/0.87 fresh(y, y, x1...xn) = u
% 3.48/0.87 C => fresh(s, t, x1...xn) = v
% 3.48/0.87 where fresh is a fresh function symbol and x1..xn are the free
% 3.48/0.87 variables of u and v.
% 3.48/0.87 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.48/0.87 input problem has no model of domain size 1).
% 3.48/0.87
% 3.48/0.87 The encoding turns the above axioms into the following unit equations and goals:
% 3.48/0.87
% 3.48/0.87 Axiom 1 (co1_6): sk1 = sk3.
% 3.48/0.87 Axiom 2 (co1_5): sk2 = sk4.
% 3.48/0.87 Axiom 3 (co1_8): segmentP(sk4, sk3) = true2.
% 3.48/0.87 Axiom 4 (co1_7): neq(sk2, nil) = true2.
% 3.48/0.87 Axiom 5 (co1_9): fresh14(X, X) = true2.
% 3.48/0.87 Axiom 6 (co1_9): fresh14(neq(sk4, nil), true2) = singletonP(sk3).
% 3.48/0.87
% 3.48/0.87 Goal 1 (co1_10): tuple2(singletonP(sk1), segmentP(sk2, sk1)) = tuple2(true2, true2).
% 3.48/0.87 Proof:
% 3.48/0.87 tuple2(singletonP(sk1), segmentP(sk2, sk1))
% 3.48/0.87 = { by axiom 1 (co1_6) }
% 3.48/0.87 tuple2(singletonP(sk3), segmentP(sk2, sk1))
% 3.48/0.88 = { by axiom 6 (co1_9) R->L }
% 3.48/0.88 tuple2(fresh14(neq(sk4, nil), true2), segmentP(sk2, sk1))
% 3.48/0.88 = { by axiom 2 (co1_5) R->L }
% 3.48/0.88 tuple2(fresh14(neq(sk2, nil), true2), segmentP(sk2, sk1))
% 3.48/0.88 = { by axiom 4 (co1_7) }
% 3.48/0.88 tuple2(fresh14(true2, true2), segmentP(sk2, sk1))
% 3.48/0.88 = { by axiom 5 (co1_9) }
% 3.48/0.88 tuple2(true2, segmentP(sk2, sk1))
% 3.48/0.88 = { by axiom 1 (co1_6) }
% 3.48/0.88 tuple2(true2, segmentP(sk2, sk3))
% 3.48/0.88 = { by axiom 2 (co1_5) }
% 3.48/0.88 tuple2(true2, segmentP(sk4, sk3))
% 3.48/0.88 = { by axiom 3 (co1_8) }
% 3.48/0.88 tuple2(true2, true2)
% 3.48/0.88 % SZS output end Proof
% 3.48/0.88
% 4.05/0.88 RESULT: Unsatisfiable (the axioms are contradictory).
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