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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

F(a, empty) → G(a, empty)
F(a, cons(x, k)) → F(cons(x, a), k)
G(cons(x, k), d) → G(k, cons(x, d))

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

G(cons(x, k), d) → G(k, cons(x, d))

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(cons(x, k), d) → G(k, cons(x, d))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)
f(x1, x2)  =  f(x1, x2)
empty  =  empty
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [RPO].
Precedence:

f₂ > g₂ > cons₁

Status:

f₂: [2,1]
g₂: [1,2]

The following usable rules [FROCOS05] were oriented:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

(9) Obligation:

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(a, cons(x, k)) → F(cons(x, a), k)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
f(x1, x2)  =  f(x1, x2)
empty  =  empty
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [RPO].
Precedence:

f₂ > g₂ > cons₁

Status:

f₂: [2,1]
g₂: [1,2]

The following usable rules [FROCOS05] were oriented:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

(14) Obligation:

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(x0, x1), x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE