YES

The TRS could be proven terminating. The proof took 20 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (7ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(+(x, s(0)))f#(x)f#(+(x, y))f#(x)
f#(s(0))f#(0)f#(+(x, y))f#(y)

Rewrite Rules

f(0)s(0)f(s(0))s(s(0))
f(s(0))*(s(s(0)), f(0))f(+(x, s(0)))+(s(s(0)), f(x))
f(+(x, y))*(f(x), f(y))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +

Strategy


The following SCCs where found

f#(+(x, s(0))) → f#(x)f#(+(x, y)) → f#(x)
f#(+(x, y)) → f#(y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(+(x, s(0)))f#(x)f#(+(x, y))f#(x)
f#(+(x, y))f#(y)

Rewrite Rules

f(0)s(0)f(s(0))s(s(0))
f(s(0))*(s(s(0)), f(0))f(+(x, s(0)))+(s(s(0)), f(x))
f(+(x, y))*(f(x), f(y))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(+(x, y))f#(x)f#(+(x, s(0)))f#(x)
f#(+(x, y))f#(y)