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:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(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:

A2(lambda1(x), y) -> A2(y, t)
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(lambda1(x), y) -> A2(x, a2(y, t))
A2(a2(x, y), z) -> A2(y, z)
A2(lambda1(x), y) -> A2(x, 1)
A2(lambda1(x), y) -> LAMBDA1(a2(x, a2(y, t)))
A2(lambda1(x), y) -> LAMBDA1(a2(x, 1))

The TRS R consists of the following rules:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(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:

A2(lambda1(x), y) -> A2(y, t)
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(lambda1(x), y) -> A2(x, a2(y, t))
A2(a2(x, y), z) -> A2(y, z)
A2(lambda1(x), y) -> A2(x, 1)
A2(lambda1(x), y) -> LAMBDA1(a2(x, a2(y, t)))
A2(lambda1(x), y) -> LAMBDA1(a2(x, 1))

The TRS R consists of the following rules:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(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 2 less nodes.

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

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

A2(lambda1(x), y) -> A2(y, t)
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(lambda1(x), y) -> A2(x, a2(y, t))
A2(a2(x, y), z) -> A2(y, z)
A2(lambda1(x), y) -> A2(x, 1)

The TRS R consists of the following rules:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A2(lambda1(x), y) -> A2(y, t)
A2(lambda1(x), y) -> A2(x, a2(y, t))
A2(lambda1(x), y) -> A2(x, 1)
The remaining pairs can at least be oriented weakly.

A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)
Used ordering: Polynomial interpretation [21]:

POL(1) = 0   
POL(A2(x1, x2)) = 2·x1 + 2·x1·x2   
POL(a2(x1, x2)) = x1 + 2·x1·x2 + x2   
POL(lambda1(x1)) = 2 + 2·x1   
POL(t) = 0   

The following usable rules [14] were oriented:

a2(x, y) -> x
lambda1(x) -> x
a2(x, y) -> y
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
a2(lambda1(x), y) -> lambda1(a2(x, 1))



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

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

A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)

The TRS R consists of the following rules:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(1) = 0   
POL(A2(x1, x2)) = 2·x1   
POL(a2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(lambda1(x1)) = 0   
POL(t) = 0   

The following usable rules [14] 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:

a2(lambda1(x), y) -> lambda1(a2(x, 1))
a2(lambda1(x), y) -> lambda1(a2(x, a2(y, t)))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(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.