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

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

IFAPPEND₃(l1, l2, false) -> APPEND₂(tl₁(l1), l2)
IFAPPEND₃(l1, l2, false) -> HD₁(l1)
APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, is_empty₁(l1))
IFAPPEND₃(l1, l2, false) -> TL₁(l1)
APPEND₂(l1, l2) -> IS_EMPTY₁(l1)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

IFAPPEND₃(l1, l2, false) -> APPEND₂(tl₁(l1), l2)
IFAPPEND₃(l1, l2, false) -> HD₁(l1)
APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, is_empty₁(l1))
IFAPPEND₃(l1, l2, false) -> TL₁(l1)
APPEND₂(l1, l2) -> IS_EMPTY₁(l1)

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

IFAPPEND₃(l1, l2, false) -> APPEND₂(tl₁(l1), l2)
APPEND₂(l1, l2) -> IFAPPEND₃(l1, l2, is_empty₁(l1))

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

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