Term Rewriting System R:
[Y, X]
minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> MINUS(X, Y)
GEQ(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.

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

The following remains to be proven:
• Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

GEQ(s(X), s(Y)) -> GEQ(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

The following remains to be proven:
• Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

GEQ(s(X), s(Y)) -> GEQ(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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

The following remains to be proven:
• Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

GEQ(s(X), s(Y)) -> GEQ(X, Y)

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

• Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y

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