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
         ↳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 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)


   R
     ↳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.


   R
     ↳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)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
EQ(x1, x2) -> EQ(x1, x2)
s(x1) -> s(x1)


   R
     ↳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.


   R
     ↳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)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MIN(x1, x2) -> x2
cons(x1, x2) -> cons(x1, x2)


   R
     ↳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.


   R
     ↳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)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
DEL(x1, x2) -> DEL(x1, x2)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳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.


   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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: 



   R
     ↳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