Termination proof

1: switching to dependency pairs

ack_in#( s( m ) , 0 )	→	u11#( ack_in( m , s( 0 ) ) )
ack_in#( s( m ) , 0 )	→	ack_in#( m , s( 0 ) )
ack_in#( s( m ) , s( n ) )	→	u21#( ack_in( s( m ) , n ) , m )
ack_in#( s( m ) , s( n ) )	→	ack_in#( s( m ) , n )
u21#( ack_out( n ) , m )	→	u22#( ack_in( m , n ) )
u21#( ack_out( n ) , m )	→	ack_in#( m , n )

1.1: dependency graph processor

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

• The 1st component contains the pair(s)

  ack_in#( s( m ) , s( n ) )	→	u21#( ack_in( s( m ) , n ) , m )
  u21#( ack_out( n ) , m )	→	ack_in#( m , n )
  ack_in#( s( m ) , 0 )	→	ack_in#( m , s( 0 ) )
  ack_in#( s( m ) , s( n ) )	→	ack_in#( s( m ) , n )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [u21# (x1, x2) ]	=	2 x1 + 1
  [ack_in (x1, x2) ]	=	2 x1 + 2 x2
  [u22 (x1) ]	=	1
  [u21 (x1, x2) ]	=	x1 + 2
  [ack_in# (x1, x2) ]	=	2 x1
  [s (x1) ]	=	x1 + 2
  [0]	=	2
  [u11 (x1) ]	=	x1
  [ack_out (x1) ]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  ack_in#( s( m ) , s( n ) )

  →

  ack_in#( s( m ) , n )

  1.1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [ack_in (x1, x2) ]	=	2 x1
  [u22 (x1) ]	=	0
  [u21 (x1, x2) ]	=	3
  [ack_in# (x1, x2) ]	=	3 x1
  [s (x1) ]	=	2 x1 + 2
  [0]	=	3
  [u11 (x1) ]	=	1
  [ack_out (x1) ]	=	0
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.1.1.1: P is empty

  All dependency pairs have been removed.