Termination w.r.t. Q proof of TRS/Cimeappend.trs

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_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

Q restricted rewrite system:
The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

hd(cons(x, l)) → x
tl(cons(x, l)) → l
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))
append(l1, l2) → ifappend(l1, l2, l1)

The TRS R 2 is

is_empty(nil) → true
is_empty(cons(x, l)) → false

The signature Sigma is {true, false, is_empty}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, nil)
ifappend(x0, x1, cons(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:

APPEND(l1, l2) → IFAPPEND(l1, l2, l1)
IFAPPEND(l1, l2, cons(x, l)) → APPEND(l, l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, nil)
ifappend(x0, x1, cons(x2, x3))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

APPEND(l1, l2) → IFAPPEND(l1, l2, l1)
IFAPPEND(l1, l2, cons(x, l)) → APPEND(l, l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, nil)
ifappend(x0, x1, cons(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

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

APPEND(l1, l2) → IFAPPEND(l1, l2, l1)
IFAPPEND(l1, l2, cons(x, l)) → APPEND(l, l2)

The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, nil)
ifappend(x0, x1, cons(x2, x3))

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.

APPEND(l1, l2) → IFAPPEND(l1, l2, l1)
IFAPPEND(l1, l2, cons(x, l)) → APPEND(l, l2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2) = APPEND(x1)
IFAPPEND(x1, x2, x3) = IFAPPEND(x3)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Precedence:

cons₂ > APPEND₁ > IFAPPEND₁

Status:

APPEND₁: multiset
IFAPPEND₁: multiset
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

The set Q consists of the following terms:

is_empty(nil)
is_empty(cons(x0, x1))
hd(cons(x0, x1))
tl(cons(x0, x1))
append(x0, x1)
ifappend(x0, x1, nil)
ifappend(x0, x1, cons(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.