Termination proof

1: switching to dependency pairs

The following set of initial dependency pairs has been identified.

+#( +( x , y ) , z ) → +#( x , +( y , z ) )
+#( +( x , y ) , z ) → +#( y , z )
+#( f( x ) , f( y ) ) → +#( x , y )
+#( f( x ) , +( f( y ) , z ) ) → +#( f( +( x , y ) ) , z )
+#( f( x ) , +( f( y ) , z ) ) → +#( x , y )

1.1: reduction pair processor

Using the following reduction pair

Linear polynomial interpretation over the naturals
[+ (x1, x2) ] = x1 + x2
[f (x1) ] = x1 + 3
[+# (x1, x2) ] = x1 + x2
[f(x1, ..., xn)] = x1 + ... + xn + 1 for all other symbols f of arity n

one remains with the following pair(s).

+#( +( x , y ) , z ) → +#( x , +( y , z ) )
+#( +( x , y ) , z ) → +#( y , z )

1.1.1: reduction pair processor

Using the following reduction pair

Linear polynomial interpretation over the naturals
[+ (x1, x2) ] = x1 + x2 + 1
[f (x1) ] = 0
[+# (x1, x2) ] = 2 x1 + x2
[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.