Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
F(s(x)) -> MINUS(s(x), g(f(x)))
F(s(x)) -> G(f(x))
F(s(x)) -> F(x)
G(s(x)) -> MINUS(s(x), f(g(x)))
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

Furthermore, R contains two SCCs.

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

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(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(MINUS(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

The following dependency pairs can be strictly oriented:

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(g(x1)) =  x1 POL(G(x1)) =  x1 POL(minus(x1, x2)) =  x1 POL(s(x1)) =  1 + x1 POL(f(x1)) =  1 + x1 POL(F(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`

Dependency Pair:

F(s(x)) -> G(f(x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Using the Dependency Graph resulted in no new DP problems.

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