Term Rewriting System R:
[x, l, l1, l2]
isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APPEND(l1, l2) -> IFAPPEND(l1, l2, isempty(l1))
APPEND(l1, l2) -> ISEMPTY(l1)
IFAPPEND(l1, l2, false) -> HD(l1)
IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
IFAPPEND(l1, l2, false) -> TL(l1)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Narrowing Transformation`

Dependency Pairs:

IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
APPEND(l1, l2) -> IFAPPEND(l1, l2, isempty(l1))

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

APPEND(l1, l2) -> IFAPPEND(l1, l2, isempty(l1))
two new Dependency Pairs are created:

APPEND(nil, l2) -> IFAPPEND(nil, l2, true)
APPEND(cons(x', l'), l2) -> IFAPPEND(cons(x', l'), l2, false)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Narrowing Transformation`

Dependency Pairs:

APPEND(cons(x', l'), l2) -> IFAPPEND(cons(x', l'), l2, false)
IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
one new Dependency Pair is created:

IFAPPEND(cons(x', l'), l2, false) -> APPEND(l', l2)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 3`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

IFAPPEND(cons(x', l'), l2, false) -> APPEND(l', l2)
APPEND(cons(x', l'), l2) -> IFAPPEND(cons(x', l'), l2, false)

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IFAPPEND(cons(x', l'), l2, false) -> APPEND(l', l2)
one new Dependency Pair is created:

IFAPPEND(cons(x', cons(x''', l''')), l2'', false) -> APPEND(cons(x''', l'''), l2'')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 4`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

IFAPPEND(cons(x', cons(x''', l''')), l2'', false) -> APPEND(cons(x''', l'''), l2'')
APPEND(cons(x', l'), l2) -> IFAPPEND(cons(x', l'), l2, false)

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

APPEND(cons(x', l'), l2) -> IFAPPEND(cons(x', l'), l2, false)
one new Dependency Pair is created:

APPEND(cons(x''', cons(x''''', l''''')), l2') -> IFAPPEND(cons(x''', cons(x''''', l''''')), l2', false)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`

Dependency Pairs:

APPEND(cons(x''', cons(x''''', l''''')), l2') -> IFAPPEND(cons(x''', cons(x''''', l''''')), l2', false)
IFAPPEND(cons(x', cons(x''', l''')), l2'', false) -> APPEND(cons(x''', l'''), l2'')

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFAPPEND(cons(x', cons(x''', l''')), l2'', false) -> APPEND(cons(x''', l'''), l2'')

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(IFAPPEND(x1, x2, x3)) =  x1 POL(cons(x1, x2)) =  1 + x2 POL(false) =  0 POL(APPEND(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 2`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`

Dependency Pair:

APPEND(cons(x''', cons(x''''', l''''')), l2') -> IFAPPEND(cons(x''', cons(x''''', l''''')), l2', false)

Rules:

isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, isempty(l1))
ifappend(l1, l2, true) -> l2
ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes