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