Termination proof

1: switching to dependency pairs

del#( .( x , .( y , z ) ) )	→	f#( =( x , y ) , x , y , z )
del#( .( x , .( y , z ) ) )	→	=#( x , y )
f#( true , x , y , z )	→	del#( .( y , z ) )
f#( false , x , y , z )	→	del#( .( y , z ) )
=#( .( x , y ) , .( u , v ) )	→	=#( x , u )
=#( .( x , y ) , .( u , v ) )	→	=#( y , v )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  f#( true , x , y , z )	→	del#( .( y , z ) )
  del#( .( x , .( y , z ) ) )	→	f#( =( x , y ) , x , y , z )
  f#( false , x , y , z )	→	del#( .( y , z ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [true]	=	2
  [v]	=	0
  [u]	=	0
  [and (x1, x2) ]	=	x1
  [f (x1, ..., x4) ]	=	2 x1 + 2 x2 + 2
  [del# (x1) ]	=	x1 + 1
  [false]	=	2
  [f# (x1, ..., x4) ]	=	x1 + 3
  [= (x1, x2) ]	=	2
  [. (x1, x2) ]	=	x1 + 1
  [nil]	=	0
  [del (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).

  del#( .( x , .( y , z ) ) )

  →

  f#( =( x , y ) , x , y , z )

  1.1.1.1: dependency graph processor

  The dependency pairs are split into 0 component(s).