TSTP Solution File: SWC262-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SWC262-1 : 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 : n021.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 20:54:41 EDT 2023
% Result : Unsatisfiable 5.33s 1.06s
% Output : Proof 5.33s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.11 % Problem : SWC262-1 : TPTP v8.1.2. Released v2.4.0.
% 0.06/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.11/0.33 % Computer : n021.cluster.edu
% 0.11/0.33 % Model : x86_64 x86_64
% 0.11/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.33 % Memory : 8042.1875MB
% 0.11/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.33 % CPULimit : 300
% 0.11/0.33 % WCLimit : 300
% 0.11/0.33 % DateTime : Mon Aug 28 18:42:43 EDT 2023
% 0.11/0.33 % CPUTime :
% 5.33/1.06 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 5.33/1.06
% 5.33/1.06 % SZS status Unsatisfiable
% 5.33/1.06
% 5.33/1.06 % SZS output start Proof
% 5.33/1.06 Take the following subset of the input axioms:
% 5.33/1.06 fof(co1_6, negated_conjecture, sk1=sk3).
% 5.33/1.06 fof(co1_7, negated_conjecture, totalorderedP(sk3)).
% 5.33/1.06 fof(co1_8, negated_conjecture, ~totalorderedP(sk1)).
% 5.33/1.06
% 5.33/1.06 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.33/1.06 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.33/1.06 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.33/1.06 fresh(y, y, x1...xn) = u
% 5.33/1.06 C => fresh(s, t, x1...xn) = v
% 5.33/1.06 where fresh is a fresh function symbol and x1..xn are the free
% 5.33/1.06 variables of u and v.
% 5.33/1.06 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.33/1.06 input problem has no model of domain size 1).
% 5.33/1.06
% 5.33/1.06 The encoding turns the above axioms into the following unit equations and goals:
% 5.33/1.06
% 5.33/1.06 Axiom 1 (co1_6): sk1 = sk3.
% 5.33/1.06 Axiom 2 (co1_7): totalorderedP(sk3) = true2.
% 5.33/1.06
% 5.33/1.06 Goal 1 (co1_8): totalorderedP(sk1) = true2.
% 5.33/1.06 Proof:
% 5.33/1.06 totalorderedP(sk1)
% 5.33/1.06 = { by axiom 1 (co1_6) }
% 5.33/1.06 totalorderedP(sk3)
% 5.33/1.06 = { by axiom 2 (co1_7) }
% 5.33/1.06 true2
% 5.33/1.06 % SZS output end Proof
% 5.33/1.06
% 5.33/1.06 RESULT: Unsatisfiable (the axioms are contradictory).
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