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:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

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:

++¹(.(x, y), z) → ++¹(y, z)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
++¹(++(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

++¹(.(x, y), z) → ++¹(y, z)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
++¹(++(x, y), z) → ++¹(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2)  =  ++¹(x1)
.(x1, x2)  =  .(x1, x2)
++(x1, x2)  =  ++(x1, x2)
nil  =  nil

Recursive path order with status [RPO].
Precedence:

++₂ > ++^1₁ > .₂
nil > .₂

Status:

++₂: multiset
++^1₁: multiset
.₂: multiset
nil: multiset

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE