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 (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| b#(r(x)) | → | b#(x) | | b#(w(x)) | → | b#(x) |
| b#(w(x)) | → | w#(b(x)) | | w#(r(x)) | → | w#(x) |
Rewrite Rules
| w(r(x)) | → | r(w(x)) | | b(r(x)) | → | r(b(x)) |
| b(w(x)) | → | w(b(x)) |
Original Signature
Termination of terms over the following signature is verified: w, b, r
Strategy
The following SCCs where found
| b#(r(x)) → b#(x) | b#(w(x)) → b#(x) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| b#(r(x)) | → | b#(x) | | b#(w(x)) | → | b#(x) |
Rewrite Rules
| w(r(x)) | → | r(w(x)) | | b(r(x)) | → | r(b(x)) |
| b(w(x)) | → | w(b(x)) |
Original Signature
Termination of terms over the following signature is verified: w, b, r
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| b#(r(x)) | → | b#(x) | | b#(w(x)) | → | b#(x) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| w(r(x)) | → | r(w(x)) | | b(r(x)) | → | r(b(x)) |
| b(w(x)) | → | w(b(x)) |
Original Signature
Termination of terms over the following signature is verified: w, b, r
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: