TSTP Solution File: COL081-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : COL081-1 : TPTP v8.1.2. Released v1.2.0.
% 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 : n014.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 : Wed Aug 30 18:32:02 EDT 2023
% Result : Unsatisfiable 0.18s 0.39s
% Output : Proof 0.18s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11 % Problem : COL081-1 : TPTP v8.1.2. Released v1.2.0.
% 0.00/0.12 % 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 : n014.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 : Sun Aug 27 04:26:02 EDT 2023
% 0.11/0.33 % CPUTime :
% 0.18/0.39 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.18/0.39
% 0.18/0.39 % SZS status Unsatisfiable
% 0.18/0.39
% 0.18/0.39 % SZS output start Proof
% 0.18/0.39 Take the following subset of the input axioms:
% 0.18/0.39 fof(abstraction, axiom, ![X, Y, Z]: apply(apply(apply(abstraction, X), Y), Z)=apply(apply(X, k(Z)), apply(Y, Z))).
% 0.18/0.39 fof(extensionality2, axiom, ![X2, Y2]: (X2=Y2 | apply(X2, n(X2, Y2))!=apply(Y2, n(X2, Y2)))).
% 0.18/0.39 fof(k_definition, axiom, ![X2, Y2]: apply(k(X2), Y2)=X2).
% 0.18/0.39 fof(prove_TRC2c, negated_conjecture, k(k(b))!=apply(abstraction, k(k(b)))).
% 0.18/0.39
% 0.18/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.18/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.18/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.18/0.39 fresh(y, y, x1...xn) = u
% 0.18/0.39 C => fresh(s, t, x1...xn) = v
% 0.18/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.18/0.39 variables of u and v.
% 0.18/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.18/0.39 input problem has no model of domain size 1).
% 0.18/0.39
% 0.18/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.18/0.39
% 0.18/0.39 Axiom 1 (k_definition): apply(k(X), Y) = X.
% 0.18/0.39 Axiom 2 (extensionality2): fresh(X, X, Y, Z) = Z.
% 0.18/0.39 Axiom 3 (abstraction): apply(apply(apply(abstraction, X), Y), Z) = apply(apply(X, k(Z)), apply(Y, Z)).
% 0.18/0.39 Axiom 4 (extensionality2): fresh(apply(X, n(X, Y)), apply(Y, n(X, Y)), X, Y) = X.
% 0.18/0.39
% 0.18/0.39 Lemma 5: fresh(apply(X, n(X, k(Y))), Y, X, k(Y)) = X.
% 0.18/0.39 Proof:
% 0.18/0.39 fresh(apply(X, n(X, k(Y))), Y, X, k(Y))
% 0.18/0.39 = { by axiom 1 (k_definition) R->L }
% 0.18/0.39 fresh(apply(X, n(X, k(Y))), apply(k(Y), n(X, k(Y))), X, k(Y))
% 0.18/0.39 = { by axiom 4 (extensionality2) }
% 0.18/0.39 X
% 0.18/0.39
% 0.18/0.39 Goal 1 (prove_TRC2c): k(k(b)) = apply(abstraction, k(k(b))).
% 0.18/0.39 Proof:
% 0.18/0.39 k(k(b))
% 0.18/0.39 = { by axiom 2 (extensionality2) R->L }
% 0.18/0.39 fresh(k(b), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by axiom 2 (extensionality2) R->L }
% 0.18/0.39 fresh(fresh(b, b, apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by axiom 1 (k_definition) R->L }
% 0.18/0.39 fresh(fresh(apply(k(b), apply(n(apply(abstraction, k(k(b))), k(k(b))), n(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)))), b, apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by axiom 1 (k_definition) R->L }
% 0.18/0.39 fresh(fresh(apply(apply(k(k(b)), k(n(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)))), apply(n(apply(abstraction, k(k(b))), k(k(b))), n(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)))), b, apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by axiom 3 (abstraction) R->L }
% 0.18/0.39 fresh(fresh(apply(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), n(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b))), b, apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b)), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by lemma 5 }
% 0.18/0.39 fresh(apply(apply(abstraction, k(k(b))), n(apply(abstraction, k(k(b))), k(k(b)))), k(b), apply(abstraction, k(k(b))), k(k(b)))
% 0.18/0.39 = { by lemma 5 }
% 0.18/0.39 apply(abstraction, k(k(b)))
% 0.18/0.39 % SZS output end Proof
% 0.18/0.39
% 0.18/0.39 RESULT: Unsatisfiable (the axioms are contradictory).
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