TSTP Solution File: GEO256+3 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : GEO256+3 : TPTP v8.1.2. Released v4.0.0.
% 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 23:28:25 EDT 2023

% Result   : Theorem 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
% 0.11/0.12  % Problem  : GEO256+3 : TPTP v8.1.2. Released v4.0.0.
% 0.14/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35  % Computer : n005.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Tue Aug 29 22:57:08 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.20/0.42  Command-line arguments: --ground-connectedness --complete-subsets
% 0.20/0.42  
% 0.20/0.42  % SZS status Theorem
% 0.20/0.42  
% 0.20/0.42  % SZS output start Proof
% 0.20/0.42  Take the following subset of the input axioms:
% 0.20/0.43    fof(a4_defns, axiom, ![X, Y]: (equally_directed_lines(X, Y) <=> ~unequally_directed_lines(X, Y))).
% 0.20/0.43    fof(a5_defns, axiom, ![X2, Y2]: (equally_directed_opposite_lines(X2, Y2) <=> ~unequally_directed_opposite_lines(X2, Y2))).
% 0.20/0.43    fof(ax10_basics, axiom, ![A, L]: ~(left_apart_point(A, L) | left_apart_point(A, reverse_line(L)))).
% 0.20/0.43    fof(ax11_basics, axiom, ![M, L2]: ~(left_convergent_lines(L2, M) | left_convergent_lines(L2, reverse_line(M)))).
% 0.20/0.43    fof(ax1_basics, axiom, ![A2]: ~distinct_points(A2, A2)).
% 0.20/0.43    fof(ax3_basics, axiom, ![L2]: ~distinct_lines(L2, L2)).
% 0.20/0.43    fof(ax5_cons_objs, axiom, ![B, A2]: (distinct_points(A2, B) => (~apart_point_and_line(A2, line_connecting(A2, B)) & ~apart_point_and_line(B, line_connecting(A2, B))))).
% 0.20/0.43    fof(ax6_cons_objs, axiom, ![L2, M2]: ((unequally_directed_lines(L2, M2) & unequally_directed_lines(L2, reverse_line(M2))) => (~apart_point_and_line(intersection_point(L2, M2), L2) & ~apart_point_and_line(intersection_point(L2, M2), M2)))).
% 0.20/0.43    fof(ax7_cons_objs, axiom, ![A2, L2]: ~apart_point_and_line(A2, parallel_through_point(L2, A2))).
% 0.20/0.43    fof(ax8_cons_objs, axiom, ![L2]: ~distinct_lines(L2, reverse_line(L2))).
% 0.20/0.43    fof(con, conjecture, ![C, D, A2, L2, B2]: ((distinct_points(A2, C) & (distinct_points(B2, C) & (incident_point_and_line(C, L2) & left_apart_point(D, L2)))) => (before_on_line(L2, A2, B2) => (before_on_line(L2, A2, C) | before_on_line(L2, C, B2))))).
% 0.20/0.43  
% 0.20/0.43  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.43  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.43  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.43    fresh(y, y, x1...xn) = u
% 0.20/0.43    C => fresh(s, t, x1...xn) = v
% 0.20/0.43  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.43  variables of u and v.
% 0.20/0.43  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.43  input problem has no model of domain size 1).
% 0.20/0.43  
% 0.20/0.43  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.43  
% 0.20/0.43  Axiom 1 (con): left_apart_point(d, l) = true2.
% 0.20/0.43  
% 0.20/0.43  Goal 1 (ax10_basics): left_apart_point(X, Y) = true2.
% 0.20/0.43  The goal is true when:
% 0.20/0.43    X = d
% 0.20/0.43    Y = l
% 0.20/0.43  
% 0.20/0.43  Proof:
% 0.20/0.43    left_apart_point(d, l)
% 0.20/0.43  = { by axiom 1 (con) }
% 0.20/0.43    true2
% 0.20/0.43  % SZS output end Proof
% 0.20/0.43  
% 0.20/0.43  RESULT: Theorem (the conjecture is true).
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