Term Rewriting System R:
[x, y]
prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

PRIME(s(s(x))) -> PRIME1(s(s(x)), s(x))
PRIME1(x, s(s(y))) -> DIVP(s(s(y)), x)
PRIME1(x, s(s(y))) -> PRIME1(x, s(y))

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`

Dependency Pair:

PRIME1(x, s(s(y))) -> PRIME1(x, s(y))

Rules:

prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

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

PRIME1(x, s(s(y))) -> PRIME1(x, s(y))
one new Dependency Pair is created:

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))

Rules:

prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

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

PRIME1(x'', s(s(s(y'')))) -> PRIME1(x'', s(s(y'')))
one new Dependency Pair is created:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 3`
`                 ↳Polynomial Ordering`

Dependency Pair:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))

Rules:

prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

The following dependency pair can be strictly oriented:

PRIME1(x'''', s(s(s(s(y''''))))) -> PRIME1(x'''', s(s(s(y''''))))

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(PRIME1(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

prime(0) -> false
prime(s(0)) -> false
prime(s(s(x))) -> prime1(s(s(x)), s(x))
prime1(x, 0) -> false
prime1(x, s(0)) -> true
prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) -> =(rem(x, y), 0)

Using the Dependency Graph resulted in no new DP problems.

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