Term Rewriting System R:
[X, L]
incr(nil) -> nil
incr(cons(X, L)) -> cons(s(X), incr(L))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
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)
ADX(cons(X, L)) -> INCR(cons(X, adx(L)))
ADX(cons(X, L)) -> ADX(L)
NATS -> ADX(zeros)
NATS -> ZEROS
ZEROS -> ZEROS

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →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))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  1 + x1 + x2 POL(INCR(x1)) =  x1

resulting in one new DP problem.
Used Argument Filtering System:
INCR(x1) -> INCR(x1)
cons(x1, x2) -> cons(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 4`
`             ↳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))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →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))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

• Dependency Pair:

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

Rules:

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

Strategy:

innermost

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →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))
adx(nil) -> nil
adx(cons(X, L)) -> incr(cons(X, adx(L)))
nats -> adx(zeros)
zeros -> cons(0, zeros)
head(cons(X, L)) -> X
tail(cons(X, L)) -> L

Strategy:

innermost

• Dependency Pair:

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

Rules:

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

Strategy:

innermost

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