(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(\, x), x) → e
app(app(\, e), x) → x
app(app(\, x), app(app(., x), y)) → y
app(app(\, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(\, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(\, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
The TRS R consists of the following rules:
app(app(\, x), x) → e
app(app(\, e), x) → x
app(app(\, x), app(app(., x), y)) → y
app(app(\, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(\, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(\, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
The TRS R consists of the following rules:
app(app(\, x), x) → e
app(app(\, e), x) → x
app(app(\, x), app(app(., x), y)) → y
app(app(\, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(\, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(\, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.