TSTP Solution File: SYN035-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN035-1 : 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 : n023.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 03:32:46 EDT 2023
% Result : Unsatisfiable 0.20s 0.38s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SYN035-1 : TPTP v8.1.2. Released v1.0.0.
% 0.11/0.13 % 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 : n023.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 19:29:11 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.38 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.38
% 0.20/0.38 % SZS status Unsatisfiable
% 0.20/0.38
% 0.20/0.38 % SZS output start Proof
% 0.20/0.39 Take the following subset of the input axioms:
% 0.20/0.39 fof(clause1, axiom, ![A, B]: p(A, B)).
% 0.20/0.39 fof(clause2, axiom, ![A2, B2]: (~p(f(A2, B2), f(A2, B2)) | (~p(B2, f(A2, B2)) | q(A2, B2)))).
% 0.20/0.39 fof(theorem, negated_conjecture, ![A3, B2]: (~q(f(A3, B2), f(A3, B2)) | (~q(A3, f(A3, B2)) | (~p(f(A3, B2), f(A3, B2)) | ~p(B2, f(A3, B2)))))).
% 0.20/0.39
% 0.20/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39 fresh(y, y, x1...xn) = u
% 0.20/0.39 C => fresh(s, t, x1...xn) = v
% 0.20/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39 variables of u and v.
% 0.20/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39 input problem has no model of domain size 1).
% 0.20/0.39
% 0.20/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39
% 0.20/0.39 Axiom 1 (clause1): p(X, Y) = true2.
% 0.20/0.39 Axiom 2 (clause2): fresh(X, X, Y, Z) = q(Y, Z).
% 0.20/0.39 Axiom 3 (clause2): fresh2(X, X, Y, Z) = true2.
% 0.20/0.39 Axiom 4 (clause2): fresh(p(f(X, Y), f(X, Y)), true2, X, Y) = fresh2(p(Y, f(X, Y)), true2, X, Y).
% 0.20/0.39
% 0.20/0.39 Lemma 5: q(X, Y) = true2.
% 0.20/0.39 Proof:
% 0.20/0.39 q(X, Y)
% 0.20/0.39 = { by axiom 2 (clause2) R->L }
% 0.20/0.39 fresh(true2, true2, X, Y)
% 0.20/0.39 = { by axiom 1 (clause1) R->L }
% 0.20/0.39 fresh(p(f(X, Y), f(X, Y)), true2, X, Y)
% 0.20/0.39 = { by axiom 4 (clause2) }
% 0.20/0.39 fresh2(p(Y, f(X, Y)), true2, X, Y)
% 0.20/0.39 = { by axiom 1 (clause1) }
% 0.20/0.39 fresh2(true2, true2, X, Y)
% 0.20/0.39 = { by axiom 3 (clause2) }
% 0.20/0.39 true2
% 0.20/0.39
% 0.20/0.39 Goal 1 (theorem): tuple(p(X, f(Y, X)), p(f(Y, X), f(Y, X)), q(Y, f(Y, X)), q(f(Y, X), f(Y, X))) = tuple(true2, true2, true2, true2).
% 0.20/0.39 The goal is true when:
% 0.20/0.39 X = Y
% 0.20/0.39 Y = X
% 0.20/0.39
% 0.20/0.39 Proof:
% 0.20/0.39 tuple(p(Y, f(X, Y)), p(f(X, Y), f(X, Y)), q(X, f(X, Y)), q(f(X, Y), f(X, Y)))
% 0.20/0.39 = { by axiom 1 (clause1) }
% 0.20/0.39 tuple(true2, p(f(X, Y), f(X, Y)), q(X, f(X, Y)), q(f(X, Y), f(X, Y)))
% 0.20/0.39 = { by axiom 1 (clause1) }
% 0.20/0.39 tuple(true2, true2, q(X, f(X, Y)), q(f(X, Y), f(X, Y)))
% 0.20/0.39 = { by lemma 5 }
% 0.20/0.39 tuple(true2, true2, true2, q(f(X, Y), f(X, Y)))
% 0.20/0.39 = { by lemma 5 }
% 0.20/0.39 tuple(true2, true2, true2, true2)
% 0.20/0.39 % SZS output end Proof
% 0.20/0.39
% 0.20/0.39 RESULT: Unsatisfiable (the axioms are contradictory).
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