Term Rewriting System R:
[x, y, z]
norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

NORM(g(x, y)) -> NORM(x)
F(x, g(y, z)) -> F(x, y)
REM(g(x, y), s(z)) -> REM(x, z)

Furthermore, R contains three SCCs.

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

Dependency Pair:

NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

The following dependency pair can be strictly oriented:

NORM(g(x, y)) -> NORM(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(g(x1, x2)) =  1 + x1 POL(NORM(x1)) =  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:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

F(x, g(y, z)) -> F(x, y)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

The following dependency pair can be strictly oriented:

F(x, g(y, z)) -> F(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(g(x1, x2)) =  1 + x1 POL(F(x1, x2)) =  x2

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:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

The following dependency pair can be strictly oriented:

REM(g(x, y), s(z)) -> REM(x, z)

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(g(x1, x2)) =  1 + x1 POL(REM(x1, x2)) =  x1 POL(s(x1)) =  0

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:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Using the Dependency Graph resulted in no new DP problems.

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