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