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.


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

APPEND2(l1, l2) -> IFAPPEND3(l1, l2, l1)
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



↳ 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))

Q is empty.
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.