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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(if_gcd, app(app(le, y), x)), app(s, x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(if_gcd, app(app(le, y), x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(le, y), x)
APP(app(gcd, app(s, x)), app(s, y)) → APP(le, y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, x), y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(minus, x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, y), x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(minus, y), x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(minus, 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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 21 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, 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.