YES
The TRS could be proven terminating. The proof took 17 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y)) | → | f#(x, s(c(s(y)))) | | f#(x, c(y)) | → | f#(x, s(f(y, y))) |
| f#(x, c(y)) | → | f#(y, y) |
Rewrite Rules
| f(x, c(y)) | → | f(x, s(f(y, y))) | | f(s(x), s(y)) | → | f(x, s(c(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: f, s, c
Strategy
The following SCCs where found
| f#(s(x), s(y)) → f#(x, s(c(s(y)))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(x, c(y)) | → | f(x, s(f(y, y))) | | f(s(x), s(y)) | → | f(x, s(c(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: f, s, c
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y)) | → | f#(x, s(c(s(y)))) |
Rewrite Rules
| f(x, c(y)) | → | f(x, s(f(y, y))) | | f(s(x), s(y)) | → | f(x, s(c(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: f, s, c
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y)) | → | f#(x, s(c(s(y)))) |