Term Rewriting System R:
[X, Y, Z]
f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(X) -> F(g(X))
F(X) -> G(X)
G(s(X)) -> G(X)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

G(s(X)) -> G(X)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

The following dependency pair can be strictly oriented:

G(s(X)) -> G(X)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(G(x1)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(SEL(x1, x2)) =  x1 POL(cons(x1, x2)) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

F(X) -> F(g(X))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Termination of R could not be shown.
Duration:
0:00 minutes