(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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)
The set Q consists of the following terms:
app(app(times, x0), 0)
app(app(times, x0), app(s, x1))
app(app(plus, x0), 0)
app(app(plus, 0), x0)
app(app(plus, x0), app(s, x1))
app(app(plus, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(times, x), app(s, y)) → APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) → APP(plus, app(app(times, x), y))
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
APP(app(plus, x), app(s, y)) → APP(s, app(app(plus, x), y))
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)
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(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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)
The set Q consists of the following terms:
app(app(times, x0), 0)
app(app(times, x0), app(s, x1))
app(app(plus, x0), 0)
app(app(plus, 0), x0)
app(app(plus, x0), app(s, x1))
app(app(plus, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 14 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, x), app(s, y)) → APP(app(plus, x), y)
The TRS R consists of the following rules:
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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)
The set Q consists of the following terms:
app(app(times, x0), 0)
app(app(times, x0), app(s, x1))
app(app(plus, x0), 0)
app(app(plus, 0), x0)
app(app(plus, x0), app(s, x1))
app(app(plus, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APP(app(times, x), app(s, y)) → APP(app(times, x), y)
The TRS R consists of the following rules:
app(app(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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)
The set Q consists of the following terms:
app(app(times, x0), 0)
app(app(times, x0), app(s, x1))
app(app(plus, x0), 0)
app(app(plus, 0), x0)
app(app(plus, x0), app(s, x1))
app(app(plus, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
(9) 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(times, x), 0) → 0
app(app(times, x), app(s, y)) → app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) → x
app(app(plus, 0), x) → x
app(app(plus, x), app(s, y)) → app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) → app(s, app(app(plus, 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)
The set Q consists of the following terms:
app(app(times, x0), 0)
app(app(times, x0), app(s, x1))
app(app(plus, x0), 0)
app(app(plus, 0), x0)
app(app(plus, x0), app(s, x1))
app(app(plus, app(s, x0)), x1)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.