Term Rewriting System R:
[X, X1, X2]
af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

AF(s(0)) -> AF(ap(s(0)))
AF(s(0)) -> AP(s(0))
MARK(f(X)) -> AF(mark(X))
MARK(f(X)) -> MARK(X)
MARK(p(X)) -> AP(mark(X))
MARK(p(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains two SCCs.

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

Dependency Pair:

AF(s(0)) -> AF(ap(s(0)))

Rules:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

The following dependency pair can be strictly oriented:

AF(s(0)) -> AF(ap(s(0)))

The following rules can be oriented:

ap(s(0)) -> 0
ap(X) -> p(X)
af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{mark, 0, ap} > p
{mark, 0, ap} > af > cons
{mark, 0, ap} > af > f
{mark, 0, ap} > af > s

resulting in one new DP problem.
Used Argument Filtering System:
AF(x1) -> AF(x1)
s(x1) -> s(x1)
ap(x1) -> ap
p(x1) -> p
af(x1) -> af(x1)
cons(x1, x2) -> cons(x1, x2)
f(x1) -> f(x1)
mark(x1) -> mark(x1)

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

Dependency Pair:

Rules:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(p(X)) -> MARK(X)
MARK(f(X)) -> MARK(X)

Rules:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

The following dependency pairs can be strictly oriented:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(f(X)) -> MARK(X)

The following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
mark > af > cons
mark > af > f
mark > af > s

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
s(x1) -> s(x1)
cons(x1, x2) -> cons(x1, x2)
p(x1) -> x1
f(x1) -> f(x1)
af(x1) -> af(x1)
ap(x1) -> x1
mark(x1) -> mark(x1)

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

Dependency Pair:

MARK(p(X)) -> MARK(X)

Rules:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

The following dependency pair can be strictly oriented:

MARK(p(X)) -> MARK(X)

The following rules can be oriented:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
mark > ap > p
mark > af > cons
mark > af > f
mark > af > 0
mark > s

resulting in one new DP problem.
Used Argument Filtering System:
MARK(x1) -> MARK(x1)
p(x1) -> p(x1)
af(x1) -> af
cons(x1, x2) -> cons(x1, x2)
f(x1) -> f
ap(x1) -> ap(x1)
s(x1) -> s(x1)
mark(x1) -> mark(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 4`
`             ↳AFS`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

af(0) -> cons(0, f(s(0)))
af(s(0)) -> af(ap(s(0)))
af(X) -> f(X)
ap(s(0)) -> 0
ap(X) -> p(X)
mark(f(X)) -> af(mark(X))
mark(p(X)) -> ap(mark(X))
mark(0) -> 0
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes