The rewrite relation of the following TRS is considered.
app(app(app(if,true),x),y) | → | x | (1) |
app(app(app(if,false),x),y) | → | y | (2) |
app(app(filter,f),nil) | → | nil | (3) |
app(app(filter,f),app(app(cons,x),xs)) | → | app(app(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))),app(app(filter,f),xs)) | (4) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))) | (5) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(cons,x),app(app(filter,f),xs)) | (7) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(if,app(f,x)) | (8) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))),app(app(filter,f),xs)) | (9) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(f,x) | (10) |
The dependency pairs are split into 1 component.
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(f,x) | (10) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
[false] | = | 1 |
[true] | = | 29393 |
[if] | = | 12079 |
[nil] | = | 24287 |
[app#(x1, x2)] | = | x2 + 0 |
[cons] | = | 60386 |
[filter] | = | 16304 |
[app(x1, x2)] | = | x1 + x2 + 31755 |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(f,x) | (10) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
app#(app(filter,f),app(app(cons,x),xs)) | → | app#(app(filter,f),xs) | (6) |
The dependency pairs are split into 0 components.