Term Rewriting System R:
[X, Y]
+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(X, s(Y)) -> +'(X, Y)
DOUBLE(X) -> +'(X, X)
F(0, s(0), X) -> F(X, double(X), X)
F(0, s(0), X) -> DOUBLE(X)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Remaining`

Dependency Pair:

+'(X, s(Y)) -> +'(X, Y)

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(X, s(Y)) -> +'(X, Y)

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:
+'(x1, x2) -> +'(x1, x2)
s(x1) -> s(x1)

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

Dependency Pair:

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

F(0, s(0), X) -> F(X, double(X), X)

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes