Term Rewriting System R:
[x, y, z]
purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

PURGE(.(x, y)) -> PURGE(remove(x, y))
PURGE(.(x, y)) -> REMOVE(x, y)
REMOVE(x, .(y, z)) -> REMOVE(x, z)

Furthermore, R contains two SCCs.

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

Dependency Pair:

REMOVE(x, .(y, z)) -> REMOVE(x, z)

Rules:

purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Strategy:

innermost

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

REMOVE(x, .(y, z)) -> REMOVE(x, z)
one new Dependency Pair is created:

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))

Rules:

purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Strategy:

innermost

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

REMOVE(x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))
one new Dependency Pair is created:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))

The transformation is resulting in one new DP problem:

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

Dependency Pair:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))

Rules:

purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REMOVE(x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', 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(REMOVE(x1, x2)) =  1 + x2 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pair:

PURGE(.(x, y)) -> PURGE(remove(x, y))

Rules:

purge(nil) -> nil
purge(.(x, y)) -> .(x, purge(remove(x, y)))
remove(x, nil) -> nil
remove(x, .(y, z)) -> if(=(x, y), remove(x, z), .(y, remove(x, z)))

Strategy:

innermost

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

PURGE(.(x, y)) -> PURGE(remove(x, y))
two new Dependency Pairs are created:

PURGE(.(x'', nil)) -> PURGE(nil)
PURGE(.(x'', .(y'', z'))) -> PURGE(if(=(x'', y''), remove(x'', z'), .(y'', remove(x'', z'))))

The transformation is resulting in no new DP problems.

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