TSTP Solution File: GEO118-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : GEO118-1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n001.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 23:27:29 EDT 2023

% Result   : Unsatisfiable 0.20s 0.50s
% 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
% 0.00/0.12  % Problem  : GEO118-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.18/0.35  % Computer : n001.cluster.edu
% 0.18/0.35  % Model    : x86_64 x86_64
% 0.18/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.18/0.35  % Memory   : 8042.1875MB
% 0.18/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.18/0.35  % CPULimit : 300
% 0.18/0.35  % WCLimit  : 300
% 0.18/0.35  % DateTime : Tue Aug 29 20:02:38 EDT 2023
% 0.18/0.35  % CPUTime  : 
% 0.20/0.50  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.50  
% 0.20/0.50  % SZS status Unsatisfiable
% 0.20/0.50  
% 0.20/0.50  % SZS output start Proof
% 0.20/0.50  Take the following subset of the input axioms:
% 0.20/0.50    fof(between_c_defn_1, axiom, ![A, B, C, D]: (~between_c(A, B, C, D) | B!=D)).
% 0.20/0.50    fof(between_o_defn_5, axiom, ![A2, B2, C2, D2]: (~ordered_by(A2, B2, C2) | (~ordered_by(A2, C2, D2) | between_o(A2, B2, C2, D2)))).
% 0.20/0.50    fof(c6_42, axiom, ![B2, A3]: (~end_point(A3, B2) | A3!=ax0_sk11(A3, B2))).
% 0.20/0.50    fof(closed_defn_31, axiom, ![B2, A3]: (~closed(A3) | ~end_point(B2, A3))).
% 0.20/0.50    fof(inner_point_defn_22, axiom, ![B2, A3]: (~inner_point(A3, B2) | ~end_point(A3, B2))).
% 0.20/0.50    fof(o3_24, axiom, ![B2, C2, D2, A2_2]: (~between_o(A2_2, B2, C2, D2) | between_c(ax2_sk4(A2_2, D2, C2, B2), B2, C2, D2))).
% 0.20/0.50    fof(theorem_4_5_133, negated_conjecture, ordered_by(sk25, sk26, sk27)).
% 0.20/0.50    fof(theorem_4_5_134, negated_conjecture, ordered_by(sk25, sk27, sk26)).
% 0.20/0.50  
% 0.20/0.50  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.50  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.50  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.50    fresh(y, y, x1...xn) = u
% 0.20/0.50    C => fresh(s, t, x1...xn) = v
% 0.20/0.50  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.50  variables of u and v.
% 0.20/0.50  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.50  input problem has no model of domain size 1).
% 0.20/0.50  
% 0.20/0.50  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.50  
% 0.20/0.50  Axiom 1 (theorem_4_5_133): ordered_by(sk25, sk26, sk27) = true2.
% 0.20/0.50  Axiom 2 (theorem_4_5_134): ordered_by(sk25, sk27, sk26) = true2.
% 0.20/0.50  Axiom 3 (between_o_defn_5): fresh61(X, X, Y, Z, W, V) = between_o(Y, Z, W, V).
% 0.20/0.50  Axiom 4 (between_o_defn_5): fresh60(X, X, Y, Z, W, V) = true2.
% 0.20/0.50  Axiom 5 (o3_24): fresh31(X, X, Y, Z, W, V) = true2.
% 0.20/0.50  Axiom 6 (between_o_defn_5): fresh61(ordered_by(X, Y, Z), true2, X, W, Y, Z) = fresh60(ordered_by(X, W, Y), true2, X, W, Y, Z).
% 0.20/0.50  Axiom 7 (o3_24): fresh31(between_o(X, Y, Z, W), true2, X, Y, Z, W) = between_c(ax2_sk4(X, W, Z, Y), Y, Z, W).
% 0.20/0.50  
% 0.20/0.50  Goal 1 (between_c_defn_1): between_c(X, Y, Z, Y) = true2.
% 0.20/0.50  The goal is true when:
% 0.20/0.50    X = ax2_sk4(sk25, sk26, sk27, sk26)
% 0.20/0.50    Y = sk26
% 0.20/0.50    Z = sk27
% 0.20/0.50  
% 0.20/0.50  Proof:
% 0.20/0.50    between_c(ax2_sk4(sk25, sk26, sk27, sk26), sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 7 (o3_24) R->L }
% 0.20/0.50    fresh31(between_o(sk25, sk26, sk27, sk26), true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 3 (between_o_defn_5) R->L }
% 0.20/0.50    fresh31(fresh61(true2, true2, sk25, sk26, sk27, sk26), true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 2 (theorem_4_5_134) R->L }
% 0.20/0.50    fresh31(fresh61(ordered_by(sk25, sk27, sk26), true2, sk25, sk26, sk27, sk26), true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 6 (between_o_defn_5) }
% 0.20/0.50    fresh31(fresh60(ordered_by(sk25, sk26, sk27), true2, sk25, sk26, sk27, sk26), true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 1 (theorem_4_5_133) }
% 0.20/0.50    fresh31(fresh60(true2, true2, sk25, sk26, sk27, sk26), true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 4 (between_o_defn_5) }
% 0.20/0.50    fresh31(true2, true2, sk25, sk26, sk27, sk26)
% 0.20/0.50  = { by axiom 5 (o3_24) }
% 0.20/0.50    true2
% 0.20/0.50  % SZS output end Proof
% 0.20/0.50  
% 0.20/0.50  RESULT: Unsatisfiable (the axioms are contradictory).
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