Termination w.r.t. Q proof of /tmp/tmpjM2vPK/z10.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

      ↳ QDPOrderProof

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

A(a(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(p(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, 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))
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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

A(a(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(a(x, y), z) → A(x, a(y, z))
A(p(x, y), z) → A(y, z)

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(1) = 0
POL(A(x₁, x₂)) = x₁ + x₂
POL(a(x₁, x₂)) = x₁ + x₂
POL(id) = 0
POL(lambda(x₁)) = 1 + x₁
POL(p(x₁, x₂)) = max(x₁, x₂)
POL(t) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPSizeChangeProof

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

A(a(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(a(x, y), z) → A(x, a(y, z))
A(p(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.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

A(a(x, y), z) → A(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
A(p(x, y), z) → A(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
A(a(x, y), z) → A(x, a(y, z))
The graph contains the following edges 1 > 1
A(p(x, y), z) → A(y, z)
The graph contains the following edges 1 > 1, 2 >= 2