TSTP Solution File: SYN014-2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN014-2 : 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 03:32:44 EDT 2023
% Result : Unsatisfiable 0.14s 0.40s
% Output : Proof 0.14s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SYN014-2 : TPTP v8.1.2. Released v1.0.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n023.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Sat Aug 26 19:55:26 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.14/0.40 Command-line arguments: --no-flatten-goal
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% 0.14/0.40 % SZS status Unsatisfiable
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% 0.14/0.40 % SZS output start Proof
% 0.14/0.40 Take the following subset of the input axioms:
% 0.14/0.40 fof(c_19, negated_conjecture, ~equalish(m, n)).
% 0.14/0.40 fof(c_20, negated_conjecture, equalish(n, k)).
% 0.14/0.40 fof(c_23, negated_conjecture, equalish(m, k)).
% 0.14/0.40 fof(symmetryish, axiom, ![X, Y]: (~equalish(X, Y) | equalish(Y, X))).
% 0.14/0.40 fof(transitivityish, axiom, ![Z, X2, Y2]: (~equalish(X2, Y2) | (~equalish(Y2, Z) | equalish(X2, Z)))).
% 0.14/0.40
% 0.14/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.14/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.14/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.14/0.40 fresh(y, y, x1...xn) = u
% 0.14/0.40 C => fresh(s, t, x1...xn) = v
% 0.14/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.14/0.40 variables of u and v.
% 0.14/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.14/0.40 input problem has no model of domain size 1).
% 0.14/0.40
% 0.14/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.14/0.40
% 0.14/0.40 Axiom 1 (c_20): equalish(n, k) = true.
% 0.14/0.40 Axiom 2 (c_23): equalish(m, k) = true.
% 0.14/0.40 Axiom 3 (transitivityish): fresh(X, X, Y, Z) = true.
% 0.14/0.40 Axiom 4 (symmetryish): fresh3(X, X, Y, Z) = true.
% 0.14/0.40 Axiom 5 (transitivityish): fresh2(X, X, Y, Z, W) = equalish(Y, W).
% 0.14/0.40 Axiom 6 (symmetryish): fresh3(equalish(X, Y), true, X, Y) = equalish(Y, X).
% 0.14/0.40 Axiom 7 (transitivityish): fresh2(equalish(X, Y), true, Z, X, Y) = fresh(equalish(Z, X), true, Z, Y).
% 0.14/0.40
% 0.14/0.40 Goal 1 (c_19): equalish(m, n) = true.
% 0.14/0.40 Proof:
% 0.14/0.40 equalish(m, n)
% 0.14/0.40 = { by axiom 5 (transitivityish) R->L }
% 0.14/0.40 fresh2(true, true, m, k, n)
% 0.14/0.40 = { by axiom 4 (symmetryish) R->L }
% 0.14/0.40 fresh2(fresh3(true, true, n, k), true, m, k, n)
% 0.14/0.40 = { by axiom 1 (c_20) R->L }
% 0.14/0.40 fresh2(fresh3(equalish(n, k), true, n, k), true, m, k, n)
% 0.14/0.40 = { by axiom 6 (symmetryish) }
% 0.14/0.40 fresh2(equalish(k, n), true, m, k, n)
% 0.14/0.40 = { by axiom 7 (transitivityish) }
% 0.14/0.40 fresh(equalish(m, k), true, m, n)
% 0.14/0.40 = { by axiom 2 (c_23) }
% 0.14/0.40 fresh(true, true, m, n)
% 0.14/0.40 = { by axiom 3 (transitivityish) }
% 0.14/0.40 true
% 0.14/0.40 % SZS output end Proof
% 0.14/0.40
% 0.14/0.40 RESULT: Unsatisfiable (the axioms are contradictory).
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