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, 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.

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

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

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

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 [FROCOS05] were oriented:

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

(6) Obligation:

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

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

(7) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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(p(x, y), z) → A(x, z)
  The graph contains the following edges 1 > 1, 2 >= 2

• A(p(x, y), z) → A(y, 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(a(x, y), z) → A(y, z)
  The graph contains the following edges 1 > 1, 2 >= 2

(8) TRUE