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`
`         ↳Argument Filtering and Ordering`

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)

The following dependency pair can be strictly oriented:

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

The following rules can be oriented:

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)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
prime > prime1 > and
prime > prime1 > true
prime > prime1 > {0, divp, false} > =
prime > prime1 > {0, divp, false} > rem
prime > prime1 > not

resulting in one new DP problem.
Used Argument Filtering System:
PRIME1(x1, x2) -> PRIME1(x1, x2)
s(x1) -> s(x1)
prime(x1) -> prime(x1)
prime1(x1, x2) -> prime1(x1, x2)
and(x1, x2) -> and(x1, x2)
not(x1) -> not(x1)
divp(x1, x2) -> divp(x1, x2)
=(x1, x2) -> =(x1, x2)
rem(x1, x2) -> rem(x1, x2)

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