YES
The TRS could be proven terminating. The proof took 27 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (13ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(x) | | g#(c(h(0))) | → | g#(d(1)) |
| f#(f(x)) | → | f#(d(f(x))) | | g#(c(1)) | → | g#(d(h(0))) |
| g#(h(x)) | → | g#(x) | | f#(f(x)) | → | f#(c(f(x))) |
Rewrite Rules
| f(f(x)) | → | f(c(f(x))) | | f(f(x)) | → | f(d(f(x))) |
| g(c(x)) | → | x | | g(d(x)) | → | x |
| g(c(h(0))) | → | g(d(1)) | | g(c(1)) | → | g(d(h(0))) |
| g(h(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h
Strategy
The following SCCs where found
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(f(x)) | → | f(c(f(x))) | | f(f(x)) | → | f(d(f(x))) |
| g(c(x)) | → | x | | g(d(x)) | → | x |
| g(c(h(0))) | → | g(d(1)) | | g(c(1)) | → | g(d(h(0))) |
| g(h(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(f(x)) | → | f(c(f(x))) | | f(f(x)) | → | f(d(f(x))) |
| g(c(x)) | → | x | | g(d(x)) | → | x |
| g(c(h(0))) | → | g(d(1)) | | g(c(1)) | → | g(d(h(0))) |
| g(h(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, d, 1, 0, c, h
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: