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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(s(x), s(y)) -> F(x, y)
G(0, x) -> G(f(x, x), x)
G(0, x) -> F(x, x)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(s(x), s(y)) -> F(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(s(s(x'')), s(s(y''))) -> F(s(x''), s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pair:

G(0, x) -> G(f(x, x), x)

Rules:

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

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

G(0, x) -> G(f(x, x), x)
two new Dependency Pairs are created:

G(0, 0) -> G(s(0), 0)
G(0, s(x'')) -> G(s(f(x'', x'')), s(x''))

The transformation is resulting in no new DP problems.

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