Term Rewriting System R:
[N, X, Y, XS]
fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FIB(N) -> SEL(N, fib1(s(0), s(0)))
FIB(N) -> FIB1(s(0), s(0))
FIB1(X, Y) -> FIB1(Y, add(X, Y))
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Furthermore, R contains three SCCs.

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

Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(N, XS)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(SEL(x1, x2)) =  x1 POL(cons(x1, x2)) =  0 POL(s(x1)) =  1 + 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`
`         ↳Remaining`

Dependency Pair:

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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(ADD(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

FIB1(X, Y) -> FIB1(Y, add(X, Y))

Rules:

fib(N) -> sel(N, fib1(s(0), s(0)))
fib1(X, Y) -> cons(X, fib1(Y, add(X, Y)))