TSTP Solution File: GEO015-3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : GEO015-3 : 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 : n031.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 23:26:50 EDT 2023
% Result : Unsatisfiable 0.20s 0.42s
% 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
% 0.12/0.13 % Problem : GEO015-3 : TPTP v8.1.2. Released v1.0.0.
% 0.12/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n031.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Tue Aug 29 21:19:41 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.42 Command-line arguments: --no-flatten-goal
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% 0.20/0.42 % SZS status Unsatisfiable
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% 0.20/0.42 % SZS output start Proof
% 0.20/0.42 Take the following subset of the input axioms:
% 0.20/0.42 fof(d1, axiom, ![V, U]: equidistant(U, V, U, V)).
% 0.20/0.42 fof(prove_symmetry, negated_conjecture, ~equidistant(w, x, u, v)).
% 0.20/0.42 fof(transitivity_for_equidistance, axiom, ![X, Y, Z, V2, W, V3]: (~equidistant(X, Y, Z, V3) | (~equidistant(X, Y, V2, W) | equidistant(Z, V3, V2, W)))).
% 0.20/0.42 fof(u_to_v_equals_w_to_x, hypothesis, equidistant(u, v, w, x)).
% 0.20/0.42
% 0.20/0.42 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.42 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.42 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.42 fresh(y, y, x1...xn) = u
% 0.20/0.42 C => fresh(s, t, x1...xn) = v
% 0.20/0.42 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.42 variables of u and v.
% 0.20/0.42 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.42 input problem has no model of domain size 1).
% 0.20/0.42
% 0.20/0.42 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.42
% 0.20/0.42 Axiom 1 (d1): equidistant(X, Y, X, Y) = true.
% 0.20/0.42 Axiom 2 (u_to_v_equals_w_to_x): equidistant(u, v, w, x) = true.
% 0.20/0.42 Axiom 3 (transitivity_for_equidistance): fresh3(X, X, Y, Z, W, V) = true.
% 0.20/0.42 Axiom 4 (transitivity_for_equidistance): fresh4(X, X, Y, Z, W, V, U, T) = equidistant(W, V, U, T).
% 0.20/0.42 Axiom 5 (transitivity_for_equidistance): fresh4(equidistant(X, Y, Z, W), true, X, Y, V, U, Z, W) = fresh3(equidistant(X, Y, V, U), true, V, U, Z, W).
% 0.20/0.42
% 0.20/0.42 Goal 1 (prove_symmetry): equidistant(w, x, u, v) = true.
% 0.20/0.42 Proof:
% 0.20/0.42 equidistant(w, x, u, v)
% 0.20/0.42 = { by axiom 4 (transitivity_for_equidistance) R->L }
% 0.20/0.42 fresh4(true, true, u, v, w, x, u, v)
% 0.20/0.42 = { by axiom 1 (d1) R->L }
% 0.20/0.42 fresh4(equidistant(u, v, u, v), true, u, v, w, x, u, v)
% 0.20/0.42 = { by axiom 5 (transitivity_for_equidistance) }
% 0.20/0.42 fresh3(equidistant(u, v, w, x), true, w, x, u, v)
% 0.20/0.42 = { by axiom 2 (u_to_v_equals_w_to_x) }
% 0.20/0.42 fresh3(true, true, w, x, u, v)
% 0.20/0.42 = { by axiom 3 (transitivity_for_equidistance) }
% 0.20/0.42 true
% 0.20/0.42 % SZS output end Proof
% 0.20/0.42
% 0.20/0.42 RESULT: Unsatisfiable (the axioms are contradictory).
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