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))

Innermost 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`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

ap(s(0)) -> 0
ap(X) -> p(X)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(a__p(x1)) =  0 POL(0) =  0 POL(A__F(x1)) =  x1 POL(s(x1)) =  1 POL(p(x1)) =  0

resulting in one new DP problem.

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

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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial 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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(s(X)) -> MARK(X)

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(MARK(x1)) =  x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  1 + x1 POL(f(x1)) =  x1 POL(p(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polynomial Ordering`

Dependency Pairs:

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(cons(X1, X2)) -> MARK(X1)

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(MARK(x1)) =  x1 POL(cons(x1, x2)) =  1 + x1 POL(f(x1)) =  x1 POL(p(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`

Dependency Pairs:

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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(MARK(x1)) =  x1 POL(f(x1)) =  x1 POL(p(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`

Dependency Pair:

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MARK(f(X)) -> MARK(X)

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(MARK(x1)) =  x1 POL(f(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Polo`
`             ...`
`               →DP Problem 7`
`                 ↳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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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