TSTP Solution File: SWC040-1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC040-1 : 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 : n013.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 20:53:33 EDT 2023
% Result : Unsatisfiable 3.62s 0.86s
% Output : Proof 3.62s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.13 % Problem : SWC040-1 : TPTP v8.1.2. Released v2.4.0.
% 0.07/0.13 % 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 : n013.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 16:21:47 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.62/0.86 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 3.62/0.86
% 3.62/0.86 % SZS status Unsatisfiable
% 3.62/0.86
% 3.62/0.86 % SZS output start Proof
% 3.62/0.86 Take the following subset of the input axioms:
% 3.62/0.86 fof(co1_5, negated_conjecture, nil=sk2).
% 3.62/0.86 fof(co1_6, negated_conjecture, sk2=sk4).
% 3.62/0.86 fof(co1_7, negated_conjecture, sk1=sk3).
% 3.62/0.86 fof(co1_8, negated_conjecture, nil!=sk1).
% 3.62/0.86 fof(co1_9, negated_conjecture, nil=sk3 | nil!=sk4).
% 3.62/0.86
% 3.62/0.86 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.62/0.86 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.62/0.86 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.62/0.86 fresh(y, y, x1...xn) = u
% 3.62/0.86 C => fresh(s, t, x1...xn) = v
% 3.62/0.86 where fresh is a fresh function symbol and x1..xn are the free
% 3.62/0.86 variables of u and v.
% 3.62/0.86 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.62/0.86 input problem has no model of domain size 1).
% 3.62/0.86
% 3.62/0.86 The encoding turns the above axioms into the following unit equations and goals:
% 3.62/0.86
% 3.62/0.86 Axiom 1 (co1_7): sk1 = sk3.
% 3.62/0.86 Axiom 2 (co1_5): nil = sk2.
% 3.62/0.86 Axiom 3 (co1_6): sk2 = sk4.
% 3.62/0.86 Axiom 4 (co1_9): fresh14(X, X) = sk3.
% 3.62/0.86 Axiom 5 (co1_9): fresh14(nil, sk4) = nil.
% 3.62/0.86
% 3.62/0.86 Goal 1 (co1_8): nil = sk1.
% 3.62/0.86 Proof:
% 3.62/0.86 nil
% 3.62/0.86 = { by axiom 5 (co1_9) R->L }
% 3.62/0.86 fresh14(nil, sk4)
% 3.62/0.86 = { by axiom 3 (co1_6) R->L }
% 3.62/0.86 fresh14(nil, sk2)
% 3.62/0.86 = { by axiom 2 (co1_5) R->L }
% 3.62/0.86 fresh14(nil, nil)
% 3.62/0.86 = { by axiom 4 (co1_9) }
% 3.62/0.86 sk3
% 3.62/0.86 = { by axiom 1 (co1_7) R->L }
% 3.62/0.86 sk1
% 3.62/0.86 % SZS output end Proof
% 3.62/0.86
% 3.62/0.86 RESULT: Unsatisfiable (the axioms are contradictory).
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