TSTP Solution File: SET158-6 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SET158-6 : TPTP v8.1.2. Bugfixed v2.1.0.
% 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 : n010.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 15:31:30 EDT 2023
% Result : Unsatisfiable 0.19s 0.55s
% 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 : SET158-6 : TPTP v8.1.2. Bugfixed v2.1.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34 % Computer : n010.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Sat Aug 26 14:06:35 EDT 2023
% 0.12/0.34 % CPUTime :
% 0.19/0.55 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.19/0.55
% 0.19/0.55 % SZS status Unsatisfiable
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% 0.19/0.55 % SZS output start Proof
% 0.19/0.55 Take the following subset of the input axioms:
% 0.19/0.55 fof(complement1, axiom, ![X, Z]: (~member(Z, complement(X)) | ~member(Z, X))).
% 0.19/0.55 fof(domain1, axiom, ![X2, Z2]: (restrict(X2, singleton(Z2), universal_class)!=null_class | ~member(Z2, domain_of(X2)))).
% 0.19/0.55 fof(prove_corollary_to_complement_axiom_1, negated_conjecture, member(y, x)).
% 0.19/0.55 fof(prove_corollary_to_complement_axiom_2, negated_conjecture, member(z, complement(x))).
% 0.19/0.55 fof(prove_corollary_to_complement_axiom_3, negated_conjecture, y=z).
% 0.19/0.55
% 0.19/0.55 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.55 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.55 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.55 fresh(y, y, x1...xn) = u
% 0.19/0.55 C => fresh(s, t, x1...xn) = v
% 0.19/0.55 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.55 variables of u and v.
% 0.19/0.55 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.55 input problem has no model of domain size 1).
% 0.19/0.55
% 0.19/0.55 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.55
% 0.19/0.55 Axiom 1 (prove_corollary_to_complement_axiom_3): y = z.
% 0.19/0.55 Axiom 2 (prove_corollary_to_complement_axiom_1): member(y, x) = true2.
% 0.19/0.55 Axiom 3 (prove_corollary_to_complement_axiom_2): member(z, complement(x)) = true2.
% 0.19/0.55
% 0.19/0.55 Goal 1 (complement1): tuple(member(X, Y), member(X, complement(Y))) = tuple(true2, true2).
% 0.19/0.55 The goal is true when:
% 0.19/0.55 X = y
% 0.19/0.55 Y = x
% 0.19/0.55
% 0.19/0.55 Proof:
% 0.19/0.55 tuple(member(y, x), member(y, complement(x)))
% 0.19/0.55 = { by axiom 2 (prove_corollary_to_complement_axiom_1) }
% 0.19/0.55 tuple(true2, member(y, complement(x)))
% 0.19/0.55 = { by axiom 1 (prove_corollary_to_complement_axiom_3) }
% 0.19/0.55 tuple(true2, member(z, complement(x)))
% 0.19/0.55 = { by axiom 3 (prove_corollary_to_complement_axiom_2) }
% 0.19/0.55 tuple(true2, true2)
% 0.19/0.55 % SZS output end Proof
% 0.19/0.55
% 0.19/0.55 RESULT: Unsatisfiable (the axioms are contradictory).
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