Termination w.r.t. Q proof of TRS/Cimeappend.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

is_empty₁(nil) -> true
is_empty₁(cons₂(x, l)) -> false
hd₁(cons₂(x, l)) -> x
tl₁(cons₂(x, l)) -> l
append₂(l1, l2) -> ifappend₃(l1, l2, l1)
ifappend₃(l1, l2, nil) -> l2
ifappend₃(l1, l2, cons₂(x, l)) -> cons₂(x, append₂(l, l2))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

is_empty₁(nil) -> true
is_empty₁(cons₂(x, l)) -> false
hd₁(cons₂(x, l)) -> x
tl₁(cons₂(x, l)) -> l
append₂(l1, l2) -> ifappend₃(l1, l2, l1)
ifappend₃(l1, l2, nil) -> l2
ifappend₃(l1, l2, cons₂(x, l)) -> cons₂(x, append₂(l, l2))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, l1)
IFAPPEND₃(l1, l2, cons₂(x, l)) -> APPEND₂(l, l2)

The TRS R consists of the following rules:

is_empty₁(nil) -> true
is_empty₁(cons₂(x, l)) -> false
hd₁(cons₂(x, l)) -> x
tl₁(cons₂(x, l)) -> l
append₂(l1, l2) -> ifappend₃(l1, l2, l1)
ifappend₃(l1, l2, nil) -> l2
ifappend₃(l1, l2, cons₂(x, l)) -> cons₂(x, append₂(l, l2))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, l1)
IFAPPEND₃(l1, l2, cons₂(x, l)) -> APPEND₂(l, l2)

The TRS R consists of the following rules:

is_empty₁(nil) -> true
is_empty₁(cons₂(x, l)) -> false
hd₁(cons₂(x, l)) -> x
tl₁(cons₂(x, l)) -> l
append₂(l1, l2) -> ifappend₃(l1, l2, l1)
ifappend₃(l1, l2, nil) -> l2
ifappend₃(l1, l2, cons₂(x, l)) -> cons₂(x, append₂(l, l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, l1)
IFAPPEND₃(l1, l2, cons₂(x, l)) -> APPEND₂(l, l2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APPEND₂(x₁, x₂)) = 2 + 3·x₁
POL(IFAPPEND₃(x₁, x₂, x₃)) = 1 + x₁ + 2·x₃
POL(cons₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

is_empty₁(nil) -> true
is_empty₁(cons₂(x, l)) -> false
hd₁(cons₂(x, l)) -> x
tl₁(cons₂(x, l)) -> l
append₂(l1, l2) -> ifappend₃(l1, l2, l1)
ifappend₃(l1, l2, nil) -> l2
ifappend₃(l1, l2, cons₂(x, l)) -> cons₂(x, append₂(l, l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.