YES
The TRS could be proven terminating. The proof took 22 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (9ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(g(X)) | → | mark#(X) | | mark#(f(X)) | → | a__f#(mark(X)) |
| mark#(f(X)) | → | mark#(X) |
Rewrite Rules
| a__f(f(a)) | → | c(f(g(f(a)))) | | mark(f(X)) | → | a__f(mark(X)) |
| mark(a) | → | a | | mark(c(X)) | → | c(X) |
| mark(g(X)) | → | g(mark(X)) | | a__f(X) | → | f(X) |
Original Signature
Termination of terms over the following signature is verified: f, g, c, a, mark, a__f
Strategy
The following SCCs where found
| mark#(g(X)) → mark#(X) | mark#(f(X)) → mark#(X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| mark#(g(X)) | → | mark#(X) | | mark#(f(X)) | → | mark#(X) |
Rewrite Rules
| a__f(f(a)) | → | c(f(g(f(a)))) | | mark(f(X)) | → | a__f(mark(X)) |
| mark(a) | → | a | | mark(c(X)) | → | c(X) |
| mark(g(X)) | → | g(mark(X)) | | a__f(X) | → | f(X) |
Original Signature
Termination of terms over the following signature is verified: f, g, c, a, mark, a__f
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| mark#(g(X)) | → | mark#(X) | | mark#(f(X)) | → | mark#(X) |