Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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)

EQ(x₁, x₂) -> EQ(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →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)

MIN(x₁, x₂) -> x₂
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and 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)

DEL(x₁, x₂) -> DEL(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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:
0:40 minutes