## TPTP Problem File: GRA030^1.p

View Solutions - Solve Problem

```%------------------------------------------------------------------------------
% File     : GRA030^1 : TPTP v8.0.0. Released v3.6.0.
% Domain   : Graph Theory
% Problem  : R(2,5) > 4
% Version  : Especial.
% English  :

%          : [Bro08] Brown (2008), Email to G. Sutcliffe
% Source   : [Bro08]
% Names    :

% Status   : Theorem
% Rating   : 0.67 v7.5.0, 0.75 v7.4.0, 0.78 v7.3.0, 0.80 v7.2.0, 0.75 v7.1.0, 0.71 v7.0.0, 0.75 v6.4.0, 0.71 v6.3.0, 0.83 v5.5.0, 0.80 v5.4.0, 0.75 v5.2.0, 1.00 v3.7.0
% Syntax   : Number of formulae    :    1 (   0 unt;   0 typ;   0 def)
%            Number of atoms       :    0 (   0 equ;   0 cnn)
%            Maximal formula atoms :    0 (   0 avg)
%            Number of connectives :  100 (  18   ~;   9   |;  22   &;  48   @)
%                                         (   0 <=>;   3  =>;   0  <=;   0 <~>)
%            Maximal formula depth :   35 (  35 avg)
%            Number of types       :    1 (   0 usr)
%            Number of type conns  :   23 (  23   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    0 (   0 usr;   0 con; --- aty)
%            Number of variables   :   15 (   0   ^  14   !;   1   ?;  15   :)
% SPC      : TH0_THM_NEQ_NAR

% Comments : If a type alpha has exactly n elements, then we can prove
%            R(k,l) > n by finding a graph (symmetric binary relation) on type
%            alpha with no k-cliques and no l-independent sets. Likewise, we
%            can prove R(k,l) <= n by proving every graph (symmetric binary
%            relation) on alpha must have a k-clique or l-independent set.
%            There is one type with 4 elements: o > o. There are two types
%            with 16 elements: o > o > o and (o > o) > o. There are two types
%            with 256 elements: o > o > o > o and o > (o > o) > o.  This means
%            we always have two formulations of R(k,l) >/<= 16 and two
%            formulations of R(k,l) >/<= 256.
%          :
%------------------------------------------------------------------------------
thf(ramsey_l_2_5_4,conjecture,
? [G: ( \$o > \$o ) > ( \$o > \$o ) > \$o] :
( ! [Xx: \$o > \$o,Xy: \$o > \$o] :
( ( G @ Xx @ Xy )
=> ( G @ Xy @ Xx ) )
& ! [Xx0: \$o > \$o,Xx1: \$o > \$o,Xp0: ( \$o > \$o ) > \$o] :
( ( ( Xp0 @ Xx0 )
& ~ ( Xp0 @ Xx1 ) )
=> ~ ( G @ Xx1 @ Xx0 ) )
& ! [Xx0: \$o > \$o,Xx1: \$o > \$o,Xx2: \$o > \$o,Xx3: \$o > \$o,Xx4: \$o > \$o,Xp0: ( \$o > \$o ) > \$o,Xp1: ( \$o > \$o ) > \$o,Xp2: ( \$o > \$o ) > \$o,Xp3: ( \$o > \$o ) > \$o] :
( ( ( Xp0 @ Xx0 )
& ~ ( Xp0 @ Xx1 )
& ~ ( Xp0 @ Xx2 )
& ~ ( Xp0 @ Xx3 )
& ~ ( Xp0 @ Xx4 )
& ~ ( Xp1 @ Xx0 )
& ( Xp1 @ Xx1 )
& ~ ( Xp1 @ Xx2 )
& ~ ( Xp1 @ Xx3 )
& ~ ( Xp1 @ Xx4 )
& ~ ( Xp2 @ Xx0 )
& ~ ( Xp2 @ Xx1 )
& ( Xp2 @ Xx2 )
& ~ ( Xp2 @ Xx3 )
& ~ ( Xp2 @ Xx4 )
& ~ ( Xp3 @ Xx0 )
& ~ ( Xp3 @ Xx1 )
& ~ ( Xp3 @ Xx2 )
& ( Xp3 @ Xx3 )
& ~ ( Xp3 @ Xx4 ) )
=> ( ( G @ Xx1 @ Xx0 )
| ( G @ Xx2 @ Xx0 )
| ( G @ Xx2 @ Xx1 )
| ( G @ Xx3 @ Xx0 )
| ( G @ Xx3 @ Xx1 )
| ( G @ Xx3 @ Xx2 )
| ( G @ Xx4 @ Xx0 )
| ( G @ Xx4 @ Xx1 )
| ( G @ Xx4 @ Xx2 )
| ( G @ Xx4 @ Xx3 ) ) ) ) ).

%------------------------------------------------------------------------------
```