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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(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 2 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(lambda(x), y) → A(y, t)
A(lambda(x), y) → A(x, a(y, t))
A(lambda(x), y) → A(x, 1)
A(a(x, y), z) → A(x, a(y, z))

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(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, a(y, t))
A(lambda(x), y) → A(x, 1)
The remaining pairs can at least be oriented weakly.

A(a(x, y), z) → A(y, z)
A(a(x, y), z) → A(x, a(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(lambda(x1)) = 4 + x_1   
POL(t) = 0   
POL(a(x1, x2)) = x_1 + x_2   
POL(A(x1, x2)) = (1/2)x_1 + (1/2)x_2   
POL(1) = 0   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(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(a(x, y), z) → A(y, z)
A(a(x, y), z) → A(x, a(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(lambda(x1)) = 4 + (4)x_1   
POL(t) = 3/2   
POL(a(x1, x2)) = 1/2 + (9/4)x_1 + x_2   
POL(A(x1, x2)) = (4)x_1   
POL(1) = 7/2   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.