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:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(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:

F(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, nil)) → G(x)
G(.(x, .(y, z))) → G(.(.(x, y), z))

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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 2 SCCs.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

[G₁, nil] > .₂

Status:

G₁: [1]
.₂: [2,1]
nil: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

[F₁, nil] > .₂

Status:

F₁: multiset
.₂: [1,2]
nil: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE