TSTP Solution File: SET239-6 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SET239-6 : TPTP v8.1.2. Bugfixed v2.1.0.
% 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 : n016.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 : Thu Aug 31 15:31:44 EDT 2023

% Result   : Unsatisfiable 0.19s 0.56s
% Output   : Proof 0.19s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SET239-6 : TPTP v8.1.2. Bugfixed v2.1.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n016.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Sat Aug 26 12:12:41 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.19/0.56  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.56  
% 0.19/0.56  % SZS status Unsatisfiable
% 0.19/0.56  
% 0.19/0.56  % SZS output start Proof
% 0.19/0.56  Take the following subset of the input axioms:
% 0.19/0.56    fof(intersection1, axiom, ![X, Y, Z]: (~member(Z, intersection(X, Y)) | member(Z, X))).
% 0.19/0.56    fof(prove_restriction_alternate_defn2_1, negated_conjecture, member(z, restrict(xr, x, y))).
% 0.19/0.56    fof(prove_restriction_alternate_defn2_2, negated_conjecture, ~member(z, xr)).
% 0.19/0.56    fof(restriction1, axiom, ![Xr, X2, Y2]: intersection(Xr, cross_product(X2, Y2))=restrict(Xr, X2, Y2)).
% 0.19/0.56  
% 0.19/0.56  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.56  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.56  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.56    fresh(y, y, x1...xn) = u
% 0.19/0.56    C => fresh(s, t, x1...xn) = v
% 0.19/0.56  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.56  variables of u and v.
% 0.19/0.56  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.56  input problem has no model of domain size 1).
% 0.19/0.56  
% 0.19/0.56  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.56  
% 0.19/0.56  Axiom 1 (restriction1): intersection(X, cross_product(Y, Z)) = restrict(X, Y, Z).
% 0.19/0.56  Axiom 2 (prove_restriction_alternate_defn2_1): member(z, restrict(xr, x, y)) = true2.
% 0.19/0.56  Axiom 3 (intersection1): fresh42(X, X, Y, Z) = true2.
% 0.19/0.56  Axiom 4 (intersection1): fresh42(member(X, intersection(Y, Z)), true2, X, Y) = member(X, Y).
% 0.19/0.56  
% 0.19/0.56  Goal 1 (prove_restriction_alternate_defn2_2): member(z, xr) = true2.
% 0.19/0.56  Proof:
% 0.19/0.56    member(z, xr)
% 0.19/0.56  = { by axiom 4 (intersection1) R->L }
% 0.19/0.56    fresh42(member(z, intersection(xr, cross_product(x, y))), true2, z, xr)
% 0.19/0.56  = { by axiom 1 (restriction1) }
% 0.19/0.56    fresh42(member(z, restrict(xr, x, y)), true2, z, xr)
% 0.19/0.56  = { by axiom 2 (prove_restriction_alternate_defn2_1) }
% 0.19/0.56    fresh42(true2, true2, z, xr)
% 0.19/0.56  = { by axiom 3 (intersection1) }
% 0.19/0.56    true2
% 0.19/0.56  % SZS output end Proof
% 0.19/0.56  
% 0.19/0.56  RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------