The rewrite relation of the following TRS is considered.
app(app(map,f),nil) | → | nil | (1) |
app(app(map,f),app(app(cons,x),xs)) | → | app(app(cons,app(f,x)),app(app(map,f),xs)) | (2) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(cons,app(f,x)) | (3) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(f,x) | (4) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(app(cons,app(f,x)),app(app(map,f),xs)) | (5) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(app(map,f),xs) | (6) |
The dependency pairs are split into 1 component.
app#(app(map,f),app(app(cons,x),xs)) | → | app#(app(map,f),xs) | (6) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(f,x) | (4) |
[nil] | = | 1 |
[app#(x1, x2)] | = | x1 + 0 |
[map] | = | 1 |
[cons] | = | 1 |
[app(x1, x2)] | = | x2 + 1 |
app(app(map,f),nil) | → | nil | (1) |
app(app(map,f),app(app(cons,x),xs)) | → | app(app(cons,app(f,x)),app(app(map,f),xs)) | (2) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(f,x) | (4) |
The dependency pairs are split into 1 component.
app#(app(map,f),app(app(cons,x),xs)) | → | app#(app(map,f),xs) | (6) |
[nil] | = | 1 |
[app#(x1, x2)] | = | x1 + x2 + 0 |
[map] | = | 1 |
[cons] | = | 1 |
[app(x1, x2)] | = | x2 + 1 |
app(app(map,f),nil) | → | nil | (1) |
app(app(map,f),app(app(cons,x),xs)) | → | app(app(cons,app(f,x)),app(app(map,f),xs)) | (2) |
app#(app(map,f),app(app(cons,x),xs)) | → | app#(app(map,f),xs) | (6) |
The dependency pairs are split into 0 components.