Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(4) Obligation:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] 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))
a(x, y) → x
a(x, y) → y
lambda(x) → x

(6) Obligation:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE