(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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(ack, 0), y) → APP(succ, y)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(ack, app(succ, x)), y) → APP(ack, x)
APP(app(ack, app(succ, x)), y) → APP(succ, 0)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(ack, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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 2 SCCs with 14 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
The TRS R consists of the following rules:
app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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.
(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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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.