YES
The TRS could be proven terminating. The proof took 16 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – 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), X) | → | f#(X, a(X)) | | f#(X, c(X)) | → | f#(s(X), X) |
Rewrite Rules
| f(s(X), X) | → | f(X, a(X)) | | f(X, c(X)) | → | f(s(X), X) |
| f(X, X) | → | c(X) |
Original Signature
Termination of terms over the following signature is verified: f, s, c, a
Strategy
The following SCCs where found
| f#(s(X), X) → f#(X, a(X)) |
| f#(X, c(X)) → f#(s(X), X) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(X, c(X)) | → | f#(s(X), X) |
Rewrite Rules
| f(s(X), X) | → | f(X, a(X)) | | f(X, c(X)) | → | f(s(X), X) |
| f(X, X) | → | c(X) |
Original Signature
Termination of terms over the following signature is verified: f, s, c, a
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(X, c(X)) | → | f#(s(X), X) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(X), X) | → | f#(X, a(X)) |
Rewrite Rules
| f(s(X), X) | → | f(X, a(X)) | | f(X, c(X)) | → | f(s(X), X) |
| f(X, X) | → | c(X) |
Original Signature
Termination of terms over the following signature is verified: f, s, c, a
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(X), X) | → | f#(X, a(X)) |