YES
The TRS could be proven terminating. The proof took 28 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (15ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(h(x, y)) | → | g#(h(y, f(x)), h(x, f(y))) | | f#(g(x, y)) | → | f#(x) |
| f#(h(x, y)) | → | f#(y) | | f#(g(x, y)) | → | g#(f(x), f(y)) |
| f#(h(x, y)) | → | f#(x) | | f#(g(x, y)) | → | f#(y) |
Rewrite Rules
| f(a) | → | b | | f(c) | → | d |
| f(g(x, y)) | → | g(f(x), f(y)) | | f(h(x, y)) | → | g(h(y, f(x)), h(x, f(y))) |
| g(x, x) | → | h(e, x) |
Original Signature
Termination of terms over the following signature is verified: f, g, d, e, b, c, a, h
Strategy
The following SCCs where found
| f#(g(x, y)) → f#(x) | f#(h(x, y)) → f#(y) |
| f#(h(x, y)) → f#(x) | f#(g(x, y)) → f#(y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(g(x, y)) | → | f#(x) | | f#(h(x, y)) | → | f#(y) |
| f#(h(x, y)) | → | f#(x) | | f#(g(x, y)) | → | f#(y) |
Rewrite Rules
| f(a) | → | b | | f(c) | → | d |
| f(g(x, y)) | → | g(f(x), f(y)) | | f(h(x, y)) | → | g(h(y, f(x)), h(x, f(y))) |
| g(x, x) | → | h(e, x) |
Original Signature
Termination of terms over the following signature is verified: f, g, d, e, b, c, a, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(h(x, y)) | → | f#(y) | | f#(g(x, y)) | → | f#(x) |
| f#(h(x, y)) | → | f#(x) | | f#(g(x, y)) | → | f#(y) |