(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(:, y)
APP(app(:, app(app(+, x), y)), z) → APP(app(+, app(app(:, x), z)), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(+, app(app(:, x), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) → APP(:, x)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(+, x), y)), z) → APP(:, y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(:, app(app(g, z), y)), app(app(+, x), a))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(:, app(app(g, z), y))
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(g, z), y)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(g, z)
APP(app(:, z), app(app(+, x), app(f, y))) → APP(app(+, x), a)
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(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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 2 SCCs with 20 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(:, app(app(:, x), y)), z) → APP(app(:, y), z)
APP(app(:, app(app(:, x), y)), z) → APP(app(:, x), app(app(:, y), z))
APP(app(:, app(app(+, x), y)), z) → APP(app(:, x), z)
APP(app(:, app(app(+, x), y)), z) → APP(app(:, y), z)
The TRS R consists of the following rules:
app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.
(6) 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(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
The TRS R consists of the following rules:
app(app(:, app(app(:, x), y)), z) → app(app(:, x), app(app(:, y), z))
app(app(:, app(app(+, x), y)), z) → app(app(+, app(app(:, x), z)), app(app(:, y), z))
app(app(:, z), app(app(+, x), app(f, y))) → app(app(:, app(app(g, z), y)), app(app(+, x), a))
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.