Term Rewriting System R:
[x, y]
top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

TOP(sent(x)) -> TOP(check(rest(x)))
TOP(sent(x)) -> CHECK(rest(x))
TOP(sent(x)) -> REST(x)
CHECK(sent(x)) -> CHECK(x)
CHECK(rest(x)) -> REST(check(x))
CHECK(rest(x)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(y)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

The following dependency pair can be strictly oriented:

CHECK(rest(x)) -> CHECK(x)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(rest(x1)) =  1 + x1 POL(cons(x1, x2)) =  x1 + x2 POL(CHECK(x1)) =  x1 POL(sent(x1)) =  x1

resulting in one new DP problem.
Used Argument Filtering System:
CHECK(x1) -> CHECK(x1)
rest(x1) -> rest(x1)
cons(x1, x2) -> cons(x1, x2)
sent(x1) -> sent(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

The following dependency pairs can be strictly oriented:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  1 + x1 + x2 POL(CHECK(x1)) =  x1 POL(sent(x1)) =  x1

resulting in one new DP problem.
Used Argument Filtering System:
CHECK(x1) -> CHECK(x1)
cons(x1, x2) -> cons(x1, x2)
sent(x1) -> sent(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳AFS`
`             ...`
`               →DP Problem 4`
`                 ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

CHECK(sent(x)) -> CHECK(x)

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

The following dependency pair can be strictly oriented:

CHECK(sent(x)) -> CHECK(x)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(CHECK(x1)) =  x1 POL(sent(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
CHECK(x1) -> CHECK(x1)
sent(x1) -> sent(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳AFS`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

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

TOP(sent(x)) -> TOP(check(rest(x)))
three new Dependency Pairs are created:

TOP(sent(x'')) -> TOP(rest(check(x'')))
TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Argument Filtering and Ordering`

Dependency Pairs:

TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))
TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(x'')) -> TOP(rest(check(x'')))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

The following dependency pair can be strictly oriented:

TOP(sent(cons(x'', y'))) -> TOP(check(sent(y')))

The following usable rules w.r.t. to the AFS can be oriented:

check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(rest(x1)) =  x1 POL(cons(x1, x2)) =  1 + x1 + x2 POL(check(x1)) =  x1 POL(nil) =  0 POL(TOP(x1)) =  1 + x1 POL(sent(x1)) =  x1

resulting in one new DP problem.
Used Argument Filtering System:
TOP(x1) -> TOP(x1)
sent(x1) -> sent(x1)
check(x1) -> check(x1)
cons(x1, x2) -> cons(x1, x2)
rest(x1) -> rest(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳AFS`
`             ...`
`               →DP Problem 7`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

TOP(sent(nil)) -> TOP(check(sent(nil)))
TOP(sent(x'')) -> TOP(rest(check(x'')))

Rules:

top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Termination of R could not be shown.
Duration:
0:01 minutes