The rewrite relation of the following TRS is considered.
| f(s(x),y,y) | → | f(y,x,s(x)) | (1) |
| f#(s(x),y,y) | → | f#(y,x,s(x)) | (2) |
The dependency pairs are split into 1 component.
| f#(s(x),y,y) | → | f#(y,x,s(x)) | (2) |
| [s(x1)] | = | x1 + 1 |
| [f(x1, x2, x3)] | = | 0 |
| [f#(x1, x2, x3)] | = | x1 + x2 + 0 |
| f#(s(x),y,y) | → | f#(y,x,s(x)) | (2) |
The dependency pairs are split into 0 components.