Term Rewriting System R:
[y, x, z]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
EQ(s(x), s(y)) -> EQ(x, y)
MINSORT(cons(x, y)) -> MIN(x, y)
MINSORT(cons(x, y)) -> MINSORT(del(min(x, y), cons(x, y)))
MINSORT(cons(x, y)) -> DEL(min(x, y), cons(x, y))
MIN(x, cons(y, z)) -> IF(le(x, y), min(x, z), min(y, z))
MIN(x, cons(y, z)) -> LE(x, y)
MIN(x, cons(y, z)) -> MIN(x, z)
MIN(x, cons(y, z)) -> MIN(y, z)
DEL(x, cons(y, z)) -> IF(eq(x, y), z, cons(y, del(x, z)))
DEL(x, cons(y, z)) -> EQ(x, y)
DEL(x, cons(y, z)) -> DEL(x, z)

Furthermore, R contains five SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

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(LE(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(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`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

EQ(s(x), s(y)) -> EQ(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EQ(s(x), s(y)) -> EQ(x, y)

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(EQ(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(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`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pairs:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MIN(x, cons(y, z)) -> MIN(y, z)
MIN(x, cons(y, z)) -> MIN(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(cons(x1, x2)) =  1 + x2 POL(MIN(x1, x2)) =  x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(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`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DEL(x, cons(y, z)) -> DEL(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(DEL(x1, x2)) =  x2 POL(cons(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 9`
`             ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Rw`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(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`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rewriting Transformation`

Dependency Pair:

MINSORT(cons(x, y)) -> MINSORT(del(min(x, y), cons(x, y)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x, y)) -> MINSORT(del(min(x, y), cons(x, y)))
one new Dependency Pair is created:

MINSORT(cons(x, y)) -> MINSORT(if(eq(min(x, y), x), y, cons(x, del(min(x, y), y))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Narrowing Transformation`

Dependency Pair:

MINSORT(cons(x, y)) -> MINSORT(if(eq(min(x, y), x), y, cons(x, del(min(x, y), y))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x, y)) -> MINSORT(if(eq(min(x, y), x), y, cons(x, del(min(x, y), y))))
six new Dependency Pairs are created:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(min(x'', nil), nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(min(x'', cons(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(min(x'', cons(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(min(x'', nil), nil))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(min(x'', nil), nil))))
one new Dependency Pair is created:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(min(x'', cons(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(min(x'', cons(y'', z')), cons(y'', z')))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', nil)))
one new Dependency Pair is created:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(min(x'', nil), x''), nil, cons(x'', del(x'', nil))))
one new Dependency Pair is created:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(min(x'', cons(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(min(x'', cons(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', del(x'', nil))))
one new Dependency Pair is created:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), cons(y'', z')))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Rewriting Transformation`

Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

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

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(min(x'', cons(y'', z')), z'))))))
one new Dependency Pair is created:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', nil)) -> MINSORT(if(eq(x'', x''), nil, cons(x'', nil)))
MINSORT(cons(x'', cons(y'', z'))) -> MINSORT(if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), x''), cons(y'', z'), cons(x'', if(eq(if(le(x'', y''), min(x'', z'), min(y'', z')), y''), z', cons(y'', del(if(le(x'', y''), min(x'', z'), min(y'', z')), z'))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
eq(0, 0) -> true
eq(0, s(y)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
minsort(nil) -> nil
minsort(cons(x, y)) -> cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) -> x
min(x, cons(y, z)) -> if(le(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, cons(y, z)) -> if(eq(x, y), z, cons(y, del(x, z)))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:22 minutes