Termination proof

1: switching to dependency pairs

admit#( x , .( u , .( v , .( w , z ) ) ) )	→	cond#( =( sum( x , u , v ) , w ) , .( u , .( v , .( w , admit( carry( x , u , v ) , z ) ) ) ) )
admit#( x , .( u , .( v , .( w , z ) ) ) )	→	admit#( carry( x , u , v ) , z )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  admit#( x , .( u , .( v , .( w , z ) ) ) )

  →

  admit#( carry( x , u , v ) , z )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [sum (x1, x2, x3) ]	=	0
  [true]	=	0
  [admit# (x1, x2) ]	=	x1
  [w]	=	2
  [admit (x1, x2) ]	=	x1 + 2 x2
  [carry (x1, x2, x3) ]	=	1
  [= (x1, x2) ]	=	0
  [. (x1, x2) ]	=	x1 + 2 x2
  [nil]	=	1
  [cond (x1, x2) ]	=	x1
  [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.