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`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(geq(x1, x2)) =  0 POL(false) =  0 POL(minus(x1, x2)) =  0 POL(MINUS(x1, x2)) =  1 + x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(div(x1, x2)) =  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`
`         ↳Polo`

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

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

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(GEQ(x1, x2)) =  1 + x1 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(geq(x1, x2)) =  0 POL(false) =  0 POL(minus(x1, x2)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(div(x1, x2)) =  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`
`         ↳Polo`

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

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

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  x2 + x3 POL(0) =  0 POL(geq(x1, x2)) =  0 POL(false) =  0 POL(DIV(x1, x2)) =  x1 POL(minus(x1, x2)) =  0 POL(true) =  0 POL(s(x1)) =  1 POL(div(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

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