Term Rewriting System R:
[y, z, x]
f(cons(nil, y)) -> y
f(cons(f(cons(nil, y)), z)) -> copy(n, y, z)
copy(0, y, z) -> f(z)
copy(s(x), y, z) -> copy(x, y, cons(f(y), z))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(cons(f(cons(nil, y)), z)) -> COPY(n, y, z)
COPY(0, y, z) -> F(z)
COPY(s(x), y, z) -> COPY(x, y, cons(f(y), z))
COPY(s(x), y, z) -> F(y)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Instantiation Transformation`

Dependency Pair:

COPY(s(x), y, z) -> COPY(x, y, cons(f(y), z))

Rules:

f(cons(nil, y)) -> y
f(cons(f(cons(nil, y)), z)) -> copy(n, y, z)
copy(0, y, z) -> f(z)
copy(s(x), y, z) -> copy(x, y, cons(f(y), z))

Strategy:

innermost

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

COPY(s(x), y, z) -> COPY(x, y, cons(f(y), z))
one new Dependency Pair is created:

COPY(s(x'), y'', cons(x'''', z'')) -> COPY(x', y'', cons(f(y''), cons(x'''', z'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Instantiation Transformation`

Dependency Pair:

COPY(s(x'), y'', cons(x'''', z'')) -> COPY(x', y'', cons(f(y''), cons(x'''', z'')))

Rules:

f(cons(nil, y)) -> y
f(cons(f(cons(nil, y)), z)) -> copy(n, y, z)
copy(0, y, z) -> f(z)
copy(s(x), y, z) -> copy(x, y, cons(f(y), z))

Strategy:

innermost

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

COPY(s(x'), y'', cons(x'''', z'')) -> COPY(x', y'', cons(f(y''), cons(x'''', z'')))
one new Dependency Pair is created:

COPY(s(x'''), y'''', cons(x''''0, cons(x'''''', z''''))) -> COPY(x''', y'''', cons(f(y''''), cons(x''''0, cons(x'''''', z''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Inst`
`             ...`
`               →DP Problem 3`
`                 ↳Polynomial Ordering`

Dependency Pair:

COPY(s(x'''), y'''', cons(x''''0, cons(x'''''', z''''))) -> COPY(x''', y'''', cons(f(y''''), cons(x''''0, cons(x'''''', z''''))))

Rules:

f(cons(nil, y)) -> y
f(cons(f(cons(nil, y)), z)) -> copy(n, y, z)
copy(0, y, z) -> f(z)
copy(s(x), y, z) -> copy(x, y, cons(f(y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

COPY(s(x'''), y'''', cons(x''''0, cons(x'''''', z''''))) -> COPY(x''', y'''', cons(f(y''''), cons(x''''0, cons(x'''''', z''''))))

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(copy(x1, x2, x3)) =  0 POL(0) =  0 POL(cons(x1, x2)) =  0 POL(COPY(x1, x2, x3)) =  x1 POL(nil) =  0 POL(n) =  0 POL(s(x1)) =  1 + x1 POL(f(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`           →DP Problem 2`
`             ↳Inst`
`             ...`
`               →DP Problem 4`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

f(cons(nil, y)) -> y
f(cons(f(cons(nil, y)), z)) -> copy(n, y, z)
copy(0, y, z) -> f(z)
copy(s(x), y, z) -> copy(x, y, cons(f(y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes