Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) -> P(s(x))
MINUS(s(x), s(y)) -> P(s(y))
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) -> MINUS(x, y)
P(s(s(x))) -> P(s(x))
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))
DIV(s(x), s(y)) -> MINUS(x, y)
DIV(plus(x, y), z) -> PLUS(div(x, z), div(y, z))
DIV(plus(x, y), z) -> DIV(x, z)
DIV(plus(x, y), z) -> DIV(y, z)
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))
PLUS(s(x), y) -> MINUS(s(x), s(0))

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pair:

PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pair:

PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pair:

PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pair:

PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

• Dependency Pairs:

DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

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