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
  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

lambda(x) → x
Used ordering:
Polynomial interpretation [25]:

POL(1) = 0   
POL(a(x1, x2)) = x1 + x2   
POL(lambda(x1)) = 2 + x1   
POL(t) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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))
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) → 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))
a(x, y) → x
a(x, y) → y

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ RuleRemovalProof

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))
a(x, y) → x
a(x, y) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPSizeChangeProof

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))
a(x, y) → x
a(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: