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)) -> 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
  ↳ Non-Overlap Check

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

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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, is_empty1(l1))
ifappend3(l1, l2, true) -> l2
ifappend3(l1, l2, false) -> cons2(hd1(l1), append2(tl1(l1), l2))

The set Q consists of the following terms:

is_empty1(nil)
is_empty1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))
append2(x0, x1)
ifappend3(x0, x1, true)
ifappend3(x0, x1, false)


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)) -> 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))

The set Q consists of the following terms:

is_empty1(nil)
is_empty1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))
append2(x0, x1)
ifappend3(x0, x1, true)
ifappend3(x0, x1, false)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)) -> 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))

The set Q consists of the following terms:

is_empty1(nil)
is_empty1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))
append2(x0, x1)
ifappend3(x0, x1, true)
ifappend3(x0, x1, false)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)) -> 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))

The set Q consists of the following terms:

is_empty1(nil)
is_empty1(cons2(x0, x1))
hd1(cons2(x0, x1))
tl1(cons2(x0, x1))
append2(x0, x1)
ifappend3(x0, x1, true)
ifappend3(x0, x1, false)

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