YES
The TRS could be proven terminating. The proof took 29 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (15ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| w#(a(a(x))) | → | u#(w(x)) | | v#(a(c(x))) | → | u#(b(d(x))) |
| v#(a(a(x))) | → | v#(x) | | w#(a(c(x))) | → | u#(b(d(x))) |
| v#(a(a(x))) | → | u#(v(x)) | | w#(a(a(x))) | → | w#(x) |
Rewrite Rules
| a(c(d(x))) | → | c(x) | | u(b(d(d(x)))) | → | b(x) |
| v(a(a(x))) | → | u(v(x)) | | v(a(c(x))) | → | u(b(d(x))) |
| v(c(x)) | → | b(x) | | w(a(a(x))) | → | u(w(x)) |
| w(a(c(x))) | → | u(b(d(x))) | | w(c(x)) | → | b(x) |
Original Signature
Termination of terms over the following signature is verified: w, v, d, u, b, c, a
Strategy
The following SCCs where found
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| a(c(d(x))) | → | c(x) | | u(b(d(d(x)))) | → | b(x) |
| v(a(a(x))) | → | u(v(x)) | | v(a(c(x))) | → | u(b(d(x))) |
| v(c(x)) | → | b(x) | | w(a(a(x))) | → | u(w(x)) |
| w(a(c(x))) | → | u(b(d(x))) | | w(c(x)) | → | b(x) |
Original Signature
Termination of terms over the following signature is verified: w, v, d, u, b, c, a
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| a(c(d(x))) | → | c(x) | | u(b(d(d(x)))) | → | b(x) |
| v(a(a(x))) | → | u(v(x)) | | v(a(c(x))) | → | u(b(d(x))) |
| v(c(x)) | → | b(x) | | w(a(a(x))) | → | u(w(x)) |
| w(a(c(x))) | → | u(b(d(x))) | | w(c(x)) | → | b(x) |
Original Signature
Termination of terms over the following signature is verified: w, v, d, u, b, c, a
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: