Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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(fact, 0) → APP(s, 0)
APP(fact, app(s, x)) → APP(app(*, app(s, x)), app(fact, app(p, app(s, x))))
APP(fact, app(s, x)) → APP(*, app(s, x))
APP(fact, app(s, x)) → APP(fact, app(p, app(s, x)))
APP(fact, app(s, x)) → APP(p, app(s, x))
APP(app(*, app(s, x)), y) → APP(app(+, app(app(*, x), y)), y)
APP(app(*, app(s, x)), y) → APP(+, app(app(*, x), y))
APP(app(*, app(s, x)), y) → APP(app(*, x), y)
APP(app(*, app(s, x)), y) → APP(*, x)
APP(app(+, x), app(s, y)) → APP(s, app(app(+, x), y))
APP(app(+, x), app(s, y)) → APP(app(+, 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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 4 SCCs with 17 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

APP(app(+, x), app(s, y)) → APP(app(+, x), y)

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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(*, app(s, x)), y) → APP(app(*, x), y)

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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(fact, app(s, x)) → APP(fact, app(p, app(s, x)))

The TRS R consists of the following rules:

app(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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.

(10) 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(p, app(s, x)) → x
app(fact, 0) → app(s, 0)
app(fact, app(s, x)) → app(app(*, app(s, x)), app(fact, app(p, app(s, x))))
app(app(*, 0), y) → 0
app(app(*, app(s, x)), y) → app(app(+, app(app(*, x), y)), y)
app(app(+, x), 0) → x
app(app(+, x), app(s, y)) → app(s, app(app(+, 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(p, app(s, x0))
app(fact, 0)
app(fact, app(s, x0))
app(app(*, 0), x0)
app(app(*, app(s, x0)), x1)
app(app(+, x0), 0)
app(app(+, x0), app(s, 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.