Term Rewriting System R:
[y, x]
leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LEQ(s(x), s(y)) -> LEQ(x, y)
-'(s(x), s(y)) -> -'(x, y)
MOD(s(x), s(y)) -> IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) -> LEQ(y, x)
MOD(s(x), s(y)) -> MOD(-(s(x), s(y)), s(y))
MOD(s(x), s(y)) -> -'(s(x), s(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:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

`   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:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

`   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:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

• Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

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