Term Rewriting System R:
[x, y, z]
msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MSORT(.(x, y)) -> MIN(x, y)
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
MSORT(.(x, y)) -> DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) -> MIN(x, z)
MIN(x, .(y, z)) -> MIN(y, z)
DEL(x, .(y, z)) -> DEL(x, z)

Furthermore, R contains three SCCs.

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

Dependency Pairs:

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

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MIN(x1, x2)) =  x2 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

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

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(DEL(x1, x2)) =  x2 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Additionally, the following usable rules for innermost can be oriented:

del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(if(x1, x2, x3)) =  0 POL(MSORT(x1)) =  x1 POL(=(x1, x2)) =  0 POL(nil) =  0 POL(min(x1, x2)) =  x1 POL(del(x1, x2)) =  0 POL(.(x1, x2)) =  1 POL(<=(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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