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))


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
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IFAPPEND3(l1, l2, cons2(x, l)) -> APPEND2(l, l2)
Used argument filtering: APPEND2(x1, x2)  =  x1
IFAPPEND3(x1, x2, x3)  =  x3
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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 contains 0 SCCs with 1 less node.