Termination w.r.t. Q proof of /tmp/tmpGmWTU9/z09.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))
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 [1,15] 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) → P(a(x, z), a(y, z))
A(p(x, y), z) → A(x, z)
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(p(x, y), z) → A(y, z)
A(lambda(x), y) → P(1, a(y, t))
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))
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

      ↳ DependencyGraphProof

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

A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(p(x, y), z) → A(x, z)
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(p(x, y), z) → A(y, z)
A(lambda(x), y) → P(1, a(y, t))
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))
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 [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ 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))
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 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(lambda(x₁)) = 1 + x₁
POL(p(x₁, x₂)) = max(x₁, x₂)
POL(t) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ 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))
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.
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