The rewrite relation of the following TRS is considered.
a(b(x)) | → | b(b(a(x))) | (1) |
a#(b(x)) | → | a#(x) | (2) |
The dependency pairs are split into 1 component.
a#(b(x)) | → | a#(x) | (2) |
[a(x1)] | = | 0 |
[b(x1)] | = | x1 + 1 |
[a#(x1)] | = | x1 + 0 |
a#(b(x)) | → | a#(x) | (2) |
The dependency pairs are split into 0 components.