YES
The TRS could be proven terminating. The proof took 41 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2)) | → | a__f#(mark(X1), X2) | | a__f#(X, X) | → | a__f#(a, b) |
| mark#(f(X1, X2)) | → | mark#(X1) | | mark#(b) | → | a__b# |
Rewrite Rules
| a__f(X, X) | → | a__f(a, b) | | a__b | → | a |
| mark(f(X1, X2)) | → | a__f(mark(X1), X2) | | mark(b) | → | a__b |
| mark(a) | → | a | | a__f(X1, X2) | → | f(X1, X2) |
| a__b | → | b |
Original Signature
Termination of terms over the following signature is verified: f, b, a, a__b, mark, a__f
Strategy
The following SCCs where found
| mark#(f(X1, X2)) → mark#(X1) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| mark#(f(X1, X2)) | → | mark#(X1) |
Rewrite Rules
| a__f(X, X) | → | a__f(a, b) | | a__b | → | a |
| mark(f(X1, X2)) | → | a__f(mark(X1), X2) | | mark(b) | → | a__b |
| mark(a) | → | a | | a__f(X1, X2) | → | f(X1, X2) |
| a__b | → | b |
Original Signature
Termination of terms over the following signature is verified: f, b, a, a__b, mark, a__f
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| mark#(f(X1, X2)) | → | mark#(X1) |