Term Rewriting System R:
[X, L]
incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

INCR(cons(X, L)) -> INCR(L)
NATS -> ZEROS
ZEROS -> ZEROS

Furthermore, R contains three SCCs.

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

Dependency Pair:

INCR(cons(X, L)) -> INCR(L)

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

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

INCR(cons(X, L)) -> INCR(L)
one new Dependency Pair is created:

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

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

INCR(cons(X, cons(X'', L''))) -> INCR(cons(X'', L''))
one new Dependency Pair is created:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

The following dependency pair can be strictly oriented:

INCR(cons(X, cons(X'''', cons(X''''', L'''')))) -> INCR(cons(X'''', cons(X''''', L'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  1 + x2 POL(INCR(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`

Dependency Pair:

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

The following remains to be proven:
• Dependency Pair:

ZEROS -> ZEROS

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

• Dependency Pair:

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

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

The following remains to be proven:
• Dependency Pair:

ZEROS -> ZEROS

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)
tail(cons(X, L)) -> L

Strategy:

innermost

• Dependency Pair:

Rules:

incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
zeros -> cons(0, zeros)