Termination proof

1: switching to dependency pairs

and#( xor( x , y ) , z )	→	xor#( and( x , z ) , and( y , z ) )
and#( xor( x , y ) , z )	→	and#( x , z )
and#( xor( x , y ) , z )	→	and#( y , z )
impl#( x , y )	→	xor#( and( x , y ) , xor( x , T ) )
impl#( x , y )	→	and#( x , y )
impl#( x , y )	→	xor#( x , T )
or#( x , y )	→	xor#( and( x , y ) , xor( x , y ) )
or#( x , y )	→	and#( x , y )
or#( x , y )	→	xor#( x , y )
equiv#( x , y )	→	xor#( x , xor( y , T ) )
equiv#( x , y )	→	xor#( y , T )
neg#( x )	→	xor#( x , T )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  and#( xor( x , y ) , z )	→	and#( y , z )
  and#( xor( x , y ) , z )	→	and#( x , z )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [neg (x1) ]	=	2 x1 + 2
  [and (x1, x2) ]	=	x1
  [impl (x1, x2) ]	=	3 x1 + 3
  [equiv (x1, x2) ]	=	2 x1 + 2 x2 + 3
  [or (x1, x2) ]	=	3 x1 + 2 x2 + 2
  [T]	=	0
  [F]	=	0
  [and# (x1, x2) ]	=	x1
  [xor (x1, x2) ]	=	x1 + x2 + 1
  [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.