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
  ↳ 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, l1)
ifappend3(l1, l2, nil) -> l2
ifappend3(l1, l2, cons2(x, l)) -> cons2(x, append2(l, 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, l1)
ifappend3(l1, l2, nil) -> l2
ifappend3(l1, l2, cons2(x, l)) -> cons2(x, append2(l, 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, nil)
ifappend3(x0, x1, cons2(x2, x3))


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))

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, nil)
ifappend3(x0, x1, cons2(x2, x3))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

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, nil)
ifappend3(x0, x1, cons2(x2, x3))

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


The following pairs can be strictly oriented and are deleted.


IFAPPEND3(l1, l2, cons2(x, l)) -> APPEND2(l, l2)
The remaining pairs can at least by weakly be oriented.

APPEND2(l1, l2) -> IFAPPEND3(l1, l2, l1)
Used ordering: Combined order from the following AFS and order.
APPEND2(x1, x2)  =  x1
IFAPPEND3(x1, x2, x3)  =  x3
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

APPEND2(l1, l2) -> IFAPPEND3(l1, l2, 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, l1)
ifappend3(l1, l2, nil) -> l2
ifappend3(l1, l2, cons2(x, l)) -> cons2(x, append2(l, 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, nil)
ifappend3(x0, x1, cons2(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.