by ttt2 (version ttt2 1.15)
The rewrite relation of the following TRS is considered.
| a(a(b(x0))) | → | c(x0) |
| a(c(x0)) | → | b(c(a(a(x0)))) |
| b(c(x0)) | → | x0 |
| a#(c(x0)) | → | a#(x0) |
| a#(c(x0)) | → | a#(a(x0)) |
| a#(c(x0)) | → | b#(c(a(a(x0)))) |
The dependency pairs are split into 1 component.
| a#(c(x0)) | → | a#(a(x0)) |
| a#(c(x0)) | → | a#(x0) |
| [a(x1)] | = | 4 · x1 + -∞ |
| [a#(x1)] | = | 0 · x1 + -∞ |
| [b(x1)] | = | -4 · x1 + 4 |
| [c(x1)] | = | 4 · x1 + 0 |
| a(a(b(x0))) | → | c(x0) |
| a(c(x0)) | → | b(c(a(a(x0)))) |
| b(c(x0)) | → | x0 |
| a#(c(x0)) | → | a#(a(x0)) |
| [a(x1)] | = | 2 · x1 + 0 |
| [a#(x1)] | = | 1/2 · x1 + 2 |
| [b(x1)] | = | 1/2 · x1 + 3/2 |
| [c(x1)] | = | 2 · x1 + 1 |
| a(a(b(x0))) | → | c(x0) |
| a(c(x0)) | → | b(c(a(a(x0)))) |
| b(c(x0)) | → | x0 |
There are no pairs anymore.