Termination proof

1: switching to dependency pairs

implies#( false , y )	→	not#( false )
implies#( x , false )	→	not#( x )
implies#( not( x ) , not( y ) )	→	implies#( y , and( x , y ) )
implies#( not( x ) , not( y ) )	→	and#( x , y )

1.1: dependency graph processor

The dependency pairs are split into 1 component(s).

• The 1st component contains the pair(s)

  implies#( not( x ) , not( y ) )

  →

  implies#( y , and( x , y ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [implies# (x1, x2) ]	=	3 x1 + 2 x2
  [and (x1, x2) ]	=	x1 + 3
  [false]	=	2
  [implies (x1, x2) ]	=	3 x1 + 2 x2 + 3
  [not (x1) ]	=	2 x1 + 2
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1: P is empty

  All dependency pairs have been removed.