(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) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) 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))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
APPEND(
x0,
x1,
x2) =
APPEND(
x1)
IFAPPEND(
x0,
x1,
x2,
x3) =
IFAPPEND(
x0,
x3)
Tags:
APPEND has argument tags [4,3,5] and root tag 1
IFAPPEND has argument tags [3,0,3,3] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
APPEND(
x1,
x2) =
APPEND
IFAPPEND(
x1,
x2,
x3) =
IFAPPEND
cons(
x1,
x2) =
cons(
x1,
x2)
Recursive path order with status [RPO].
Quasi-Precedence:
cons2 > [APPEND, IFAPPEND]
Status:
APPEND: multiset
IFAPPEND: multiset
cons2: multiset
The following usable rules [FROCOS05] were oriented:
none
(4) 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))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(6) TRUE