Termination w.r.t. Q proof of TRS/Zantemaz10.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))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, 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))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, 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(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))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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))
a(id, x) → x
a(1, id) → 1
a(t, id) → t
a(1, p(x, y)) → x
a(t, p(x, y)) → y

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