The rewrite relation of the following TRS is considered.
f(nil) | → | nil | (1) |
f(.(nil,y)) | → | .(nil,f(y)) | (2) |
f(.(.(x,y),z)) | → | f(.(x,.(y,z))) | (3) |
g(nil) | → | nil | (4) |
g(.(x,nil)) | → | .(g(x),nil) | (5) |
g(.(x,.(y,z))) | → | g(.(.(x,y),z)) | (6) |
g#(.(x,.(y,z))) | → | g#(.(.(x,y),z)) | (7) |
g#(.(x,nil)) | → | g#(x) | (8) |
f#(.(nil,y)) | → | f#(y) | (9) |
f#(.(.(x,y),z)) | → | f#(.(x,.(y,z))) | (10) |
The dependency pairs are split into 2 components.
f#(.(.(x,y),z)) | → | f#(.(x,.(y,z))) | (10) |
f#(.(nil,y)) | → | f#(y) | (9) |
[f(x1)] | = | 0 |
[nil] | = | 1 |
[f#(x1)] | = | x1 + 0 |
[g#(x1)] | = | 0 |
[.(x1, x2)] | = | x1 + x2 + 28958 |
[g(x1)] | = | 0 |
f#(.(nil,y)) | → | f#(y) | (9) |
The dependency pairs are split into 1 component.
f#(.(.(x,y),z)) | → | f#(.(x,.(y,z))) | (10) |
[f(x1)] | = | 0 |
[nil] | = | 1 |
[f#(x1)] | = | x1 + 0 |
[g#(x1)] | = | 0 |
[.(x1, x2)] | = | x1 + 32286 |
[g(x1)] | = | 0 |
f#(.(.(x,y),z)) | → | f#(.(x,.(y,z))) | (10) |
The dependency pairs are split into 0 components.
g#(.(x,nil)) | → | g#(x) | (8) |
g#(.(x,.(y,z))) | → | g#(.(.(x,y),z)) | (7) |
[f(x1)] | = | 0 |
[nil] | = | 1 |
[f#(x1)] | = | 0 |
[g#(x1)] | = | x1 + 0 |
[.(x1, x2)] | = | x1 + x2 + 1424 |
[g(x1)] | = | 0 |
g#(.(x,nil)) | → | g#(x) | (8) |
The dependency pairs are split into 1 component.
g#(.(x,.(y,z))) | → | g#(.(.(x,y),z)) | (7) |
[f(x1)] | = | 0 |
[nil] | = | 1 |
[f#(x1)] | = | 0 |
[g#(x1)] | = | x1 + 0 |
[.(x1, x2)] | = | x2 + 4421 |
[g(x1)] | = | 0 |
g#(.(x,.(y,z))) | → | g#(.(.(x,y),z)) | (7) |
The dependency pairs are split into 0 components.