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 Order [17,21] with Interpretation:

POL( IFAPPEND3(x1, ..., x3) ) = 2x2 + x3 + 3


POL( cons2(x1, x2) ) = x1 + 2x2 + 1


POL( APPEND2(x1, x2) ) = x1 + 2x2 + 3



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.