YES

The TRS could be proven terminating. The proof took 19 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (3ms).
 | – Problem 2 was processed with processor SubtermCriterion (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

a__f#(f(a))a__f#(g(f(a)))mark#(f(X))a__f#(mark(X))
mark#(f(X))mark#(X)

Rewrite Rules

a__f(f(a))a__f(g(f(a)))mark(f(X))a__f(mark(X))
mark(a)amark(g(X))g(X)
a__f(X)f(X)

Original Signature

Termination of terms over the following signature is verified: f, g, a, mark, a__f

Strategy


The following SCCs where found

mark#(f(X)) → mark#(X)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

mark#(f(X))mark#(X)

Rewrite Rules

a__f(f(a))a__f(g(f(a)))mark(f(X))a__f(mark(X))
mark(a)amark(g(X))g(X)
a__f(X)f(X)

Original Signature

Termination of terms over the following signature is verified: f, g, a, mark, a__f

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

mark#(f(X))mark#(X)