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.