Term Rewriting System R:
[YS, X, XS, Y, L]
app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

APP(cons(X, XS), YS) -> APP(XS, YS)
FROM(X) -> FROM(s(X))
ZWADR(cons(X, XS), cons(Y, YS)) -> APP(Y, cons(X, nil))
PREFIX(L) -> PREFIX(L)

Furthermore, R contains four SCCs.

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

Dependency Pair:

APP(cons(X, XS), YS) -> APP(XS, YS)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

APP(cons(X, XS), YS) -> APP(XS, YS)
one new Dependency Pair is created:

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

APP(cons(X, cons(X'', XS'')), YS'') -> APP(cons(X'', XS''), YS'')
one new Dependency Pair is created:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(cons(X, cons(X'''', cons(X''''', XS''''))), YS'''') -> APP(cons(X'''', cons(X''''', XS'''')), YS'''')

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(APP(x1, x2)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

FROM(X) -> FROM(s(X))
one new Dependency Pair is created:

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`           →DP Problem 8`
`             ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

FROM(s(X'')) -> FROM(s(s(X'')))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

FROM(s(X'')) -> FROM(s(s(X'')))
one new Dependency Pair is created:

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`           →DP Problem 8`
`             ↳Inst`
`             ...`
`               →DP Problem 9`
`                 ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
one new Dependency Pair is created:

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`           →DP Problem 8`
`             ↳Inst`
`             ...`
`               →DP Problem 10`
`                 ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
one new Dependency Pair is created:

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`           →DP Problem 8`
`             ↳Inst`
`             ...`
`               →DP Problem 11`
`                 ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

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

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
one new Dependency Pair is created:

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Inst`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))

Strategy:

innermost

• Dependency Pair:

PREFIX(L) -> PREFIX(L)

Rules:

app(nil, YS) -> YS
app(cons(X, XS), YS) -> cons(X, app(XS, YS))
from(X) -> cons(X, from(s(X)))