Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] 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, is_empty, false}

(2) Obligation:

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))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
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)

Recursive path order with status [RPO].
Precedence:

cons₂ > APPEND₁ > IFAPPEND₁

Status:

cons₂: multiset
APPEND₁: multiset
IFAPPEND₁: multiset

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE