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:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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(x, :(y, z))
:¹(:(x, y), z) → :¹(y, z)
:¹(+(x, y), z) → :¹(x, z)
:¹(+(x, y), z) → :¹(y, z)
:¹(z, +(x, f(y))) → :¹(g(z, y), +(x, a))

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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 1 less node.

(4) 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(x, z)
:¹(+(x, y), z) → :¹(y, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
:¹(x0, x1, x2)  =  :¹(x0)

Tags:
:¹ has argument tags [0,2,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
:¹(x1, x2)  =  :¹(x1, x2)
:(x1, x2)  =  :(x1, x2)
+(x1, x2)  =  +(x1, x2)
f(x1)  =  f
g(x1, x2)  =  x1
a  =  a

Recursive path order with status [RPO].
Quasi-Precedence:

[:^1₂, :₂, f, a] > +₂

Status:

:^1₂: [1,2]
:₂: [1,2]
+₂: multiset
f: multiset
a: multiset

The following usable rules [FROCOS05] were oriented:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

(6) Obligation:

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

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE