Termination proof

1: switching to dependency pairs

The following set of initial dependency pairs has been identified.

f#( f( x , y , z ) , u , f( x , y , v ) ) → f#( x , y , f( z , u , v ) )
f#( f( x , y , z ) , u , f( x , y , v ) ) → f#( z , u , v )

1.1: reduction pair processor

Using the following reduction pair

Linear polynomial interpretation over the naturals
[f (x1, x2, x3) ] = 2 x1 + 2 x2 + 3
[f# (x1, x2, x3) ] = x1 + x2
[g (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: P is empty

All dependency pairs have been removed.