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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(s(0)) -> F(0)
F(+(x, s(0))) -> F(x)
F(+(x, y)) -> F(x)
F(+(x, y)) -> F(y)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`

Dependency Pairs:

F(+(x, y)) -> F(y)
F(+(x, y)) -> F(x)
F(+(x, s(0))) -> F(x)

Rules:

f(0) -> s(0)
f(s(0)) -> s(s(0))
f(s(0)) -> *(s(s(0)), f(0))
f(+(x, s(0))) -> +(s(s(0)), f(x))
f(+(x, y)) -> *(f(x), f(y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(+(x, y)) -> F(y)
F(+(x, y)) -> F(x)
F(+(x, s(0))) -> F(x)

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
+(x1, x2) -> +(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 2`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

f(0) -> s(0)
f(s(0)) -> s(s(0))
f(s(0)) -> *(s(s(0)), f(0))
f(+(x, s(0))) -> +(s(s(0)), f(x))
f(+(x, y)) -> *(f(x), f(y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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