Term Rewriting System R:
[m, n, x, k]
eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(s(n), s(m)) -> EQ(n, m)
LE(s(n), s(m)) -> LE(n, m)
MIN(cons(n, cons(m, x))) -> IFMIN(le(n, m), cons(n, cons(m, x)))
MIN(cons(n, cons(m, x))) -> LE(n, m)
IFMIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
IFMIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
REPLACE(n, m, cons(k, x)) -> IFREPLACE(eq(n, k), n, m, cons(k, x))
REPLACE(n, m, cons(k, x)) -> EQ(n, k)
IFREPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
SORT(cons(n, x)) -> MIN(cons(n, x))
SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))
SORT(cons(n, x)) -> REPLACE(min(cons(n, x)), n, x)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

EQ(s(n), s(m)) -> EQ(n, m)


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

EQ(s(n), s(m)) -> EQ(n, m)


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 6
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


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
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

LE(s(n), s(m)) -> LE(n, m)


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

LE(s(n), s(m)) -> LE(n, m)


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 2
AFS
           →DP Problem 7
Dependency Graph
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


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
Narrowing Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
REPLACE(n, m, cons(k, x)) -> IFREPLACE(eq(n, k), n, m, cons(k, x))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

REPLACE(n, m, cons(k, x)) -> IFREPLACE(eq(n, k), n, m, cons(k, x))
four new Dependency Pairs are created:

REPLACE(0, m, cons(0, x)) -> IFREPLACE(true, 0, m, cons(0, x))
REPLACE(0, m, cons(s(m''), x)) -> IFREPLACE(false, 0, m, cons(s(m''), x))
REPLACE(s(n''), m, cons(0, x)) -> IFREPLACE(false, s(n''), m, cons(0, x))
REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
REPLACE(s(n''), m, cons(0, x)) -> IFREPLACE(false, s(n''), m, cons(0, x))
REPLACE(0, m, cons(s(m''), x)) -> IFREPLACE(false, 0, m, cons(s(m''), x))
IFREPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

IFREPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
three new Dependency Pairs are created:

IFREPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')
IFREPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')
IFREPLACE(false, s(n''''), m'', cons(s(m''''), x')) -> REPLACE(s(n''''), m'', x')

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')
REPLACE(s(n''), m, cons(0, x)) -> IFREPLACE(false, s(n''), m, cons(0, x))
IFREPLACE(false, s(n''''), m'', cons(s(m''''), x')) -> REPLACE(s(n''''), m'', x')
REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

IFREPLACE(false, s(n''''), m'', cons(0, x'')) -> REPLACE(s(n''''), m'', x'')
two new Dependency Pairs are created:

IFREPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
IFREPLACE(false, s(n'''''), m''0, cons(0, cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 11
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, s(n''''), m'', cons(s(m''''), x')) -> REPLACE(s(n''''), m'', x')
REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
IFREPLACE(false, s(n'''''), m''0, cons(0, cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))
IFREPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n''), m, cons(0, x)) -> IFREPLACE(false, s(n''), m, cons(0, x))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

REPLACE(s(n''), m, cons(0, x)) -> IFREPLACE(false, s(n''), m, cons(0, x))
two new Dependency Pairs are created:

REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
REPLACE(s(n'''), m', cons(0, cons(s(m''''''), x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(s(m''''''), x''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 13
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, s(n'''''), m''0, cons(0, cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))
REPLACE(s(n'''), m', cons(0, cons(s(m''''''), x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(s(m''''''), x''''')))
IFREPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
IFREPLACE(false, s(n''''), m'', cons(s(m''''), x')) -> REPLACE(s(n''''), m'', x')


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

IFREPLACE(false, s(n''''), m'', cons(s(m''''), x')) -> REPLACE(s(n''''), m'', x')
three new Dependency Pairs are created:

IFREPLACE(false, s(n'''''), m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))
IFREPLACE(false, s(n''''''), m'''', cons(s(m''''), cons(0, cons(0, x''''''')))) -> REPLACE(s(n''''''), m'''', cons(0, cons(0, x''''''')))
IFREPLACE(false, s(n''''''), m'''', cons(s(m''''), cons(0, cons(s(m''''''''), x''''''')))) -> REPLACE(s(n''''''), m'''', cons(0, cons(s(m''''''''), x''''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 15
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, s(n''''''), m'''', cons(s(m''''), cons(0, cons(s(m''''''''), x''''''')))) -> REPLACE(s(n''''''), m'''', cons(0, cons(s(m''''''''), x''''''')))
REPLACE(s(n'''), m', cons(0, cons(s(m''''''), x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(s(m''''''), x''''')))
IFREPLACE(false, s(n'''''), m''', cons(0, cons(0, x'''))) -> REPLACE(s(n'''''), m''', cons(0, x'''))
REPLACE(s(n'''), m', cons(0, cons(0, x'''''))) -> IFREPLACE(false, s(n'''), m', cons(0, cons(0, x''''')))
IFREPLACE(false, s(n''''''), m'''', cons(s(m''''), cons(0, cons(0, x''''''')))) -> REPLACE(s(n''''''), m'''', cons(0, cons(0, x''''''')))
IFREPLACE(false, s(n'''''), m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))
REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
IFREPLACE(false, s(n'''''), m''0, cons(0, cons(s(m''''), x'''))) -> REPLACE(s(n'''''), m''0, cons(s(m''''), x'''))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

REPLACE(s(n''), m, cons(s(m''), x)) -> IFREPLACE(eq(n'', m''), s(n''), m, cons(s(m''), x))
three new Dependency Pairs are created:

REPLACE(s(n'''), m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IFREPLACE(eq(n''', m'''), s(n'''), m', cons(s(m'''), cons(s(m'''''''), x''''')))
REPLACE(s(n'''), m', cons(s(m'''), cons(0, cons(0, x''''''''')))) -> IFREPLACE(eq(n''', m'''), s(n'''), m', cons(s(m'''), cons(0, cons(0, x'''''''''))))
REPLACE(s(n'''), m', cons(s(m'''), cons(0, cons(s(m''''''''''), x''''''''')))) -> IFREPLACE(eq(n''', m'''), s(n'''), m', cons(s(m'''), cons(0, cons(s(m''''''''''), x'''''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 10
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')
REPLACE(0, m, cons(s(m''), x)) -> IFREPLACE(false, 0, m, cons(s(m''), x))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

IFREPLACE(false, 0, m'', cons(s(m''''), x'')) -> REPLACE(0, m'', x'')
one new Dependency Pair is created:

IFREPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 12
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

IFREPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))
REPLACE(0, m, cons(s(m''), x)) -> IFREPLACE(false, 0, m, cons(s(m''), x))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




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

REPLACE(0, m, cons(s(m''), x)) -> IFREPLACE(false, 0, m, cons(s(m''), x))
one new Dependency Pair is created:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IFREPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 14
Argument Filtering and Ordering
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IFREPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))
IFREPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFREPLACE(false, 0, m''0, cons(s(m''''), cons(s(m''''), x'''))) -> REPLACE(0, m''0, cons(s(m''''), x'''))


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:
REPLACE(x1, x2, x3) -> x3
cons(x1, x2) -> cons(x1, x2)
IFREPLACE(x1, x2, x3, x4) -> x4
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 8
Inst
             ...
               →DP Problem 17
Dependency Graph
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

REPLACE(0, m', cons(s(m'''), cons(s(m'''''''), x'''''))) -> IFREPLACE(false, 0, m', cons(s(m'''), cons(s(m'''''''), x''''')))


Rules:


eq(0, 0) -> true
eq(0, s(m)) -> false
eq(s(n), 0) -> false
eq(s(n), s(m)) -> eq(n, m)
le(0, m) -> true
le(s(n), 0) -> false
le(s(n), s(m)) -> le(n, m)
min(cons(0, nil)) -> 0
min(cons(s(n), nil)) -> s(n)
min(cons(n, cons(m, x))) -> ifmin(le(n, m), cons(n, cons(m, x)))
ifmin(true, cons(n, cons(m, x))) -> min(cons(n, x))
ifmin(false, cons(n, cons(m, x))) -> min(cons(m, x))
replace(n, m, nil) -> nil
replace(n, m, cons(k, x)) -> ifreplace(eq(n, k), n, m, cons(k, x))
ifreplace(true, n, m, cons(k, x)) -> cons(m, x)
ifreplace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
sort(nil) -> nil
sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))


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
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:04 minutes