YES
The TRS could be proven terminating. The proof took 25 ms.
The following DP Processors were used
Problem 1 was processed with processor SubtermCriterion (1ms).
| – Problem 2 was processed with processor DependencyGraph (2ms).
| | – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(g(x), g(y)) | → | h#(g(y), x, g(y)) | | h#(g(x), y, z) | → | h#(x, y, z) |
| f#(g(x), a) | → | f#(x, g(a)) | | h#(g(x), y, z) | → | f#(y, h(x, y, z)) |
Rewrite Rules
| f(a, g(y)) | → | g(g(y)) | | f(g(x), a) | → | f(x, g(a)) |
| f(g(x), g(y)) | → | h(g(y), x, g(y)) | | h(g(x), y, z) | → | f(y, h(x, y, z)) |
| h(a, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(g(x), g(y)) | → | h#(g(y), x, g(y)) | | f#(g(x), a) | → | f#(x, g(a)) |
Problem 2: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| h#(g(x), y, z) | → | h#(x, y, z) | | h#(g(x), y, z) | → | f#(y, h(x, y, z)) |
Rewrite Rules
| f(a, g(y)) | → | g(g(y)) | | f(g(x), a) | → | f(x, g(a)) |
| f(g(x), g(y)) | → | h(g(y), x, g(y)) | | h(g(x), y, z) | → | f(y, h(x, y, z)) |
| h(a, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
The following SCCs where found
| h#(g(x), y, z) → h#(x, y, z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| h#(g(x), y, z) | → | h#(x, y, z) |
Rewrite Rules
| f(a, g(y)) | → | g(g(y)) | | f(g(x), a) | → | f(x, g(a)) |
| f(g(x), g(y)) | → | h(g(y), x, g(y)) | | h(g(x), y, z) | → | f(y, h(x, y, z)) |
| h(a, y, z) | → | z |
Original Signature
Termination of terms over the following signature is verified: f, g, a, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| h#(g(x), y, z) | → | h#(x, y, z) |