Termination w.r.t. Q proof of TRS/Zantemaz09.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(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A(p(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → A(y, t)
A(lambda(x), y) → LAMBDA(a(x, p(1, a(y, t))))
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(lambda(x), y) → P(1, a(y, t))

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

A(p(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → A(y, t)
A(lambda(x), y) → LAMBDA(a(x, p(1, a(y, t))))
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(lambda(x), y) → P(1, a(y, t))

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A(p(x, y), z) → A(x, z)
A(p(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → LAMBDA(a(x, p(1, a(y, t))))
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(lambda(x), y) → P(1, a(y, t))

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

A(p(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(x, p(1, a(y, t)))
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y

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