Termination proof

1: switching to dependency pairs

max#( N( L( s( x ) ) , L( s( y ) ) ) )	→	max#( N( L( x ) , L( y ) ) )
max#( N( L( x ) , N( y , z ) ) )	→	max#( N( L( x ) , L( max( N( y , z ) ) ) ) )
max#( N( L( x ) , N( y , z ) ) )	→	max#( N( y , z ) )

1.1: dependency graph processor

The dependency pairs are split into 2 component(s).

• The 1st component contains the pair(s)

  max#( N( L( x ) , N( y , z ) ) )

  →

  max#( N( y , z ) )

  1.1.1: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [N (x1, x2) ]	=	3 x1 + 3 x2 + 1
  [L (x1) ]	=	x1
  [max (x1) ]	=	x1
  [0]	=	0
  [s (x1) ]	=	0
  [max# (x1) ]	=	2 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.

• The 2nd component contains the pair(s)

  max#( N( L( s( x ) ) , L( s( y ) ) ) )

  →

  max#( N( L( x ) , L( y ) ) )

  1.1.2: reduction pair processor

  Using the following reduction pair

  Linear polynomial interpretation over the naturals

  [N (x1, x2) ]	=	x1
  [L (x1) ]	=	x1
  [max (x1) ]	=	x1
  [s (x1) ]	=	2 x1 + 1
  [0]	=	0
  [max# (x1) ]	=	x1
  [f(x1, ..., xn)]	=	x1 + ... + xn + 1	for all other symbols f of arity n

  one remains with the following pair(s).

  none

  1.1.2.1: P is empty

  All dependency pairs have been removed.