The rewrite relation of the following TRS is considered.
| a(b(x1)) | → | b(b(a(x1))) | (1) |
| b(c(x1)) | → | c(b(b(x1))) | (2) |
| c(a(x1)) | → | a(c(c(x1))) | (3) |
| b#(c(x1)) | → | b#(b(x1)) | (4) |
| c#(a(x1)) | → | c#(x1) | (5) |
| a#(b(x1)) | → | a#(x1) | (6) |
| a#(b(x1)) | → | b#(b(a(x1))) | (7) |
| c#(a(x1)) | → | c#(c(x1)) | (8) |
| b#(c(x1)) | → | b#(x1) | (9) |
| b#(c(x1)) | → | c#(b(b(x1))) | (10) |
| a#(b(x1)) | → | b#(a(x1)) | (11) |
| c#(a(x1)) | → | a#(c(c(x1))) | (12) |
The dependency pairs are split into 1 component.
| c#(a(x1)) | → | a#(c(c(x1))) | (12) |
| a#(b(x1)) | → | b#(a(x1)) | (11) |
| b#(c(x1)) | → | c#(b(b(x1))) | (10) |
| b#(c(x1)) | → | b#(x1) | (9) |
| a#(b(x1)) | → | a#(x1) | (6) |
| c#(a(x1)) | → | c#(x1) | (5) |
| c#(a(x1)) | → | c#(c(x1)) | (8) |
| a#(b(x1)) | → | b#(b(a(x1))) | (7) |
| b#(c(x1)) | → | b#(b(x1)) | (4) |
| [a(x1)] | = | 1 |
| [b(x1)] | = | x1 + 0 |
| [c(x1)] | = | x1 + 4 |
| [c#(x1)] | = | 3 |
| [a#(x1)] | = | 2 |
| [b#(x1)] | = | x1 + 0 |
| a(b(x1)) | → | b(b(a(x1))) | (1) |
| c(a(x1)) | → | a(c(c(x1))) | (3) |
| b(c(x1)) | → | c(b(b(x1))) | (2) |
| c#(a(x1)) | → | a#(c(c(x1))) | (12) |
| a#(b(x1)) | → | b#(a(x1)) | (11) |
| b#(c(x1)) | → | c#(b(b(x1))) | (10) |
| b#(c(x1)) | → | b#(x1) | (9) |
| a#(b(x1)) | → | b#(b(a(x1))) | (7) |
| b#(c(x1)) | → | b#(b(x1)) | (4) |
The dependency pairs are split into 2 components.
| a#(b(x1)) | → | a#(x1) | (6) |
| [a(x1)] | = | 0 |
| [b(x1)] | = | x1 + 1 |
| [c(x1)] | = | 0 |
| [c#(x1)] | = | 0 |
| [a#(x1)] | = | x1 + 0 |
| [b#(x1)] | = | 0 |
| a#(b(x1)) | → | a#(x1) | (6) |
The dependency pairs are split into 0 components.
| c#(a(x1)) | → | c#(x1) | (5) |
| c#(a(x1)) | → | c#(c(x1)) | (8) |
| [a(x1)] | = | x1 + 1 |
| [b(x1)] | = | 41063 |
| [c(x1)] | = | x1 + 0 |
| [c#(x1)] | = | x1 + 3 |
| [a#(x1)] | = | 2 |
| [b#(x1)] | = | x1 + 0 |
| a(b(x1)) | → | b(b(a(x1))) | (1) |
| c(a(x1)) | → | a(c(c(x1))) | (3) |
| b(c(x1)) | → | c(b(b(x1))) | (2) |
| c#(a(x1)) | → | c#(x1) | (5) |
| c#(a(x1)) | → | c#(c(x1)) | (8) |
The dependency pairs are split into 0 components.