TSTP Solution File: GEO019-3 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : GEO019-3 : 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 : n005.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 23:26:52 EDT 2023
% Result : Unsatisfiable 0.13s 0.40s
% Output : Proof 0.13s
% 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 : GEO019-3 : 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
% 0.13/0.33 % Computer : n005.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Tue Aug 29 21:23:07 EDT 2023
% 0.13/0.33 % CPUTime :
% 0.13/0.40 Command-line arguments: --flatten
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% 0.13/0.40 % SZS status Unsatisfiable
% 0.13/0.40
% 0.13/0.40 % SZS output start Proof
% 0.13/0.40 Take the following subset of the input axioms:
% 0.13/0.40 fof(d2, axiom, ![X, V, W, U]: (~equidistant(U, V, W, X) | equidistant(W, X, U, V))).
% 0.13/0.40 fof(d3, axiom, ![V2, X2, W2, U2]: (~equidistant(U2, V2, W2, X2) | equidistant(V2, U2, W2, X2))).
% 0.13/0.40 fof(prove_symmetry, negated_conjecture, ~equidistant(w, x, v, u)).
% 0.13/0.40 fof(u_to_v_equals_w_to_x, hypothesis, equidistant(u, v, w, x)).
% 0.13/0.40
% 0.13/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.40 fresh(y, y, x1...xn) = u
% 0.13/0.40 C => fresh(s, t, x1...xn) = v
% 0.13/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.40 variables of u and v.
% 0.13/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.40 input problem has no model of domain size 1).
% 0.13/0.40
% 0.13/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.40
% 0.13/0.40 Axiom 1 (u_to_v_equals_w_to_x): equidistant(u, v, w, x) = true.
% 0.13/0.40 Axiom 2 (d2): fresh10(X, X, Y, Z, W, V) = true.
% 0.13/0.40 Axiom 3 (d3): fresh9(X, X, Y, Z, W, V) = true.
% 0.13/0.40 Axiom 4 (d2): fresh10(equidistant(X, Y, Z, W), true, X, Y, Z, W) = equidistant(Z, W, X, Y).
% 0.13/0.40 Axiom 5 (d3): fresh9(equidistant(X, Y, Z, W), true, X, Y, Z, W) = equidistant(Y, X, Z, W).
% 0.13/0.40
% 0.13/0.40 Goal 1 (prove_symmetry): equidistant(w, x, v, u) = true.
% 0.13/0.40 Proof:
% 0.13/0.40 equidistant(w, x, v, u)
% 0.13/0.40 = { by axiom 4 (d2) R->L }
% 0.13/0.40 fresh10(equidistant(v, u, w, x), true, v, u, w, x)
% 0.13/0.40 = { by axiom 1 (u_to_v_equals_w_to_x) R->L }
% 0.13/0.40 fresh10(equidistant(v, u, w, x), equidistant(u, v, w, x), v, u, w, x)
% 0.13/0.40 = { by axiom 5 (d3) R->L }
% 0.13/0.40 fresh10(fresh9(equidistant(u, v, w, x), true, u, v, w, x), equidistant(u, v, w, x), v, u, w, x)
% 0.13/0.40 = { by axiom 1 (u_to_v_equals_w_to_x) R->L }
% 0.13/0.40 fresh10(fresh9(equidistant(u, v, w, x), equidistant(u, v, w, x), u, v, w, x), equidistant(u, v, w, x), v, u, w, x)
% 0.13/0.40 = { by axiom 3 (d3) }
% 0.13/0.40 fresh10(true, equidistant(u, v, w, x), v, u, w, x)
% 0.13/0.40 = { by axiom 1 (u_to_v_equals_w_to_x) R->L }
% 0.13/0.40 fresh10(equidistant(u, v, w, x), equidistant(u, v, w, x), v, u, w, x)
% 0.13/0.40 = { by axiom 2 (d2) }
% 0.13/0.40 true
% 0.13/0.40 % SZS output end Proof
% 0.13/0.40
% 0.13/0.40 RESULT: Unsatisfiable (the axioms are contradictory).
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