Termination w.r.t. Q proof of /tmp/tmptU2NEi/thiemann39.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(div2, x)
A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(not, f), x) → A(not2, a(f, x))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(filter, a(not, a(divides, x)))
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))
A(a(a(if, true), x), xs) → A(cons, x)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(div2, x), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(a(a(div2, x), y), 0) → A(divides, x)
A(sieve, a(a(cons, x), xs)) → A(not, a(divides, x))
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(divides, a(s, x)), a(s, y)) → A(div2, x)
A(sieve, a(a(cons, x), xs)) → A(divides, x)
A(a(divides, a(s, x)), a(s, y)) → A(a(div2, x), a(s, y))
A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(div2, x)
A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(not, f), x) → A(not2, a(f, x))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(filter, a(not, a(divides, x)))
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))
A(a(a(if, true), x), xs) → A(cons, x)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(div2, x), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(a(a(div2, x), y), 0) → A(divides, x)
A(sieve, a(a(cons, x), xs)) → A(not, a(divides, x))
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(divides, a(s, x)), a(s, y)) → A(div2, x)
A(sieve, a(a(cons, x), xs)) → A(divides, x)
A(a(divides, a(s, x)), a(s, y)) → A(a(div2, x), a(s, y))
A(a(not, f), x) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

A(a(a(div2, x), y), 0) → A(a(divides, x), y)
A(a(divides, a(s, x)), a(s, y)) → A(a(a(div2, x), a(s, y)), y)
A(a(a(div2, a(s, x)), y), a(s, z)) → A(a(a(div2, x), y), z)

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

A(sieve, a(a(cons, x), xs)) → A(a(filter, a(not, a(divides, x))), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(not, f), x) → A(f, x)
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(sieve, a(a(cons, x), xs)) → A(sieve, a(a(filter, a(not, a(divides, x))), xs))

The TRS R consists of the following rules:

a(a(divides, 0), a(s, y)) → true
a(a(divides, a(s, x)), a(s, y)) → a(a(a(div2, x), a(s, y)), y)
a(a(a(div2, x), y), 0) → a(a(divides, x), y)
a(a(a(div2, 0), y), a(s, z)) → false
a(a(a(div2, a(s, x)), y), a(s, z)) → a(a(a(div2, x), y), z)
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), x) → a(not2, a(f, x))
a(not2, true) → false
a(not2, false) → true
a(sieve, nil) → nil
a(sieve, a(a(cons, x), xs)) → a(a(cons, x), a(sieve, a(a(filter, a(not, a(divides, x))), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.