Term Rewriting System R:
[x, y, z]
double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

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

Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

The following dependency pair can be strictly oriented:

DOUBLE(s(x)) -> DOUBLE(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(DOUBLE(x1)) =  x1 POL(s(x1)) =  1 + 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:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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:

HALF(s(s(x))) -> HALF(x)

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(HALF(x1)) =  1 + x1 POL(s(x1)) =  1 + 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:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

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:

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

Rules:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(-'(x1, x2)) =  x1 POL(s(x1)) =  1 + 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:

double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Using the Dependency Graph resulted in no new DP problems.

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