Term Rewriting System R:
[x, y, n, m]
eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(s(x), s(y)) -> EQ(x, y)
LE(s(x), s(y)) -> LE(x, y)
APP(add(n, x), y) -> APP(x, y)
MIN(add(n, add(m, x))) -> IFMIN(le(n, m), add(n, add(m, x)))
MIN(add(n, add(m, x))) -> LE(n, m)
IFMIN(true, add(n, add(m, x))) -> MIN(add(n, x))
IFMIN(false, add(n, add(m, x))) -> MIN(add(m, x))
RM(n, add(m, x)) -> IFRM(eq(n, m), n, add(m, x))
RM(n, add(m, x)) -> EQ(n, m)
IFRM(true, n, add(m, x)) -> RM(n, x)
IFRM(false, n, add(m, x)) -> RM(n, x)
MINSORT(add(n, x), y) -> IFMINSORT(eq(n, min(add(n, x))), add(n, x), y)
MINSORT(add(n, x), y) -> EQ(n, min(add(n, x)))
MINSORT(add(n, x), y) -> MIN(add(n, x))
IFMINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
IFMINSORT(true, add(n, x), y) -> APP(rm(n, x), y)
IFMINSORT(true, add(n, x), y) -> RM(n, x)
IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))

Furthermore, R contains six SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:

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


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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 7
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:

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


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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 2
AFS
           →DP Problem 8
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:

APP(add(n, x), y) -> APP(x, y)


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(add(n, x), y) -> APP(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:
APP(x1, x2) -> APP(x1, x2)
add(x1, x2) -> add(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
           →DP Problem 9
Dependency Graph
       →DP Problem 4
Nar
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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
Narrowing Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(false, n, add(m, x)) -> RM(n, x)
IFRM(true, n, add(m, x)) -> RM(n, x)
RM(n, add(m, x)) -> IFRM(eq(n, m), n, add(m, x))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

RM(n, add(m, x)) -> IFRM(eq(n, m), n, add(m, x))
four new Dependency Pairs are created:

RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))

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
Nar
           →DP Problem 10
Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
IFRM(true, n, add(m, x)) -> RM(n, x)
RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))
IFRM(false, n, add(m, x)) -> RM(n, x)


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(true, n, add(m, x)) -> RM(n, x)
two new Dependency Pairs are created:

IFRM(true, 0, add(0, x'')) -> RM(0, x'')
IFRM(true, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 11
Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
IFRM(true, 0, add(0, x'')) -> RM(0, x'')
RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))
IFRM(false, n, add(m, x)) -> RM(n, x)
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(false, n, add(m, x)) -> RM(n, x)
three new Dependency Pairs are created:

IFRM(false, 0, add(s(x''''), x'')) -> RM(0, x'')
IFRM(false, s(x''''), add(0, x'')) -> RM(s(x''''), x'')
IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 12
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
IFRM(false, s(x''''), add(0, x'')) -> RM(s(x''''), x'')
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
IFRM(true, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(true, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
two new Dependency Pairs are created:

IFRM(true, s(x''''''), add(s(y'''), add(0, x0'))) -> RM(s(x''''''), add(0, x0'))
IFRM(true, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 14
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))
IFRM(true, s(x''''''), add(s(y'''), add(0, x0'))) -> RM(s(x''''''), add(0, x0'))
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
IFRM(false, s(x''''), add(0, x'')) -> RM(s(x''''), x'')
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(false, s(x''''), add(0, x'')) -> RM(s(x''''), x'')
two new Dependency Pairs are created:

IFRM(false, s(x'''''), add(0, add(0, x'0))) -> RM(s(x'''''), add(0, x'0))
IFRM(false, s(x'''''), add(0, add(s(y'''), x'0))) -> RM(s(x'''''), add(s(y'''), x'0))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 16
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, s(x''''''), add(s(y'''), add(0, x0'))) -> RM(s(x''''''), add(0, x0'))
IFRM(false, s(x'''''), add(0, add(s(y'''), x'0))) -> RM(s(x'''''), add(s(y'''), x'0))
IFRM(false, s(x'''''), add(0, add(0, x'0))) -> RM(s(x'''''), add(0, x'0))
RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
IFRM(true, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

RM(s(x''), add(0, x)) -> IFRM(false, s(x''), add(0, x))
two new Dependency Pairs are created:

RM(s(x'''), add(0, add(0, x'0''))) -> IFRM(false, s(x'''), add(0, add(0, x'0'')))
RM(s(x'''), add(0, add(s(y'''''), x'0''))) -> IFRM(false, s(x'''), add(0, add(s(y'''''), x'0'')))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 18
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))
IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
IFRM(false, s(x'''''), add(0, add(s(y'''), x'0))) -> RM(s(x'''''), add(s(y'''), x'0))
RM(s(x'''), add(0, add(s(y'''''), x'0''))) -> IFRM(false, s(x'''), add(0, add(s(y'''''), x'0'')))
IFRM(false, s(x'''''), add(0, add(0, x'0))) -> RM(s(x'''''), add(0, x'0))
RM(s(x'''), add(0, add(0, x'0''))) -> IFRM(false, s(x'''), add(0, add(0, x'0'')))
IFRM(true, s(x''''''), add(s(y'''), add(0, x0'))) -> RM(s(x''''''), add(0, x0'))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(false, s(x'''''), add(s(y'''), x')) -> RM(s(x'''''), x')
three new Dependency Pairs are created:

IFRM(false, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))
IFRM(false, s(x''''''), add(s(y'''), add(0, add(0, x'0'''')))) -> RM(s(x''''''), add(0, add(0, x'0'''')))
IFRM(false, s(x''''''), add(s(y'''), add(0, add(s(y'''''''), x'0'''')))) -> RM(s(x''''''), add(0, add(s(y'''''''), x'0'''')))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 20
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(false, s(x''''''), add(s(y'''), add(0, add(s(y'''''''), x'0'''')))) -> RM(s(x''''''), add(0, add(s(y'''''''), x'0'''')))
IFRM(false, s(x''''''), add(s(y'''), add(0, add(0, x'0'''')))) -> RM(s(x''''''), add(0, add(0, x'0'''')))
IFRM(false, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))
IFRM(false, s(x'''''), add(0, add(s(y'''), x'0))) -> RM(s(x'''''), add(s(y'''), x'0))
RM(s(x'''), add(0, add(s(y'''''), x'0''))) -> IFRM(false, s(x'''), add(0, add(s(y'''''), x'0'')))
IFRM(false, s(x'''''), add(0, add(0, x'0))) -> RM(s(x'''''), add(0, x'0))
RM(s(x'''), add(0, add(0, x'0''))) -> IFRM(false, s(x'''), add(0, add(0, x'0'')))
IFRM(true, s(x''''''), add(s(y'''), add(0, x0'))) -> RM(s(x''''''), add(0, x0'))
RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
IFRM(true, s(x''''''), add(s(y'''), add(s(y'''), x0'))) -> RM(s(x''''''), add(s(y'''), x0'))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

RM(s(x''), add(s(y'), x)) -> IFRM(eq(x'', y'), s(x''), add(s(y'), x))
four new Dependency Pairs are created:

RM(s(x''''), add(s(y''), add(0, x0'''))) -> IFRM(eq(x'''', y''), s(x''''), add(s(y''), add(0, x0''')))
RM(s(x''''), add(s(y''), add(s(y''''''), x0'''))) -> IFRM(eq(x'''', y''), s(x''''), add(s(y''), add(s(y''''''), x0''')))
RM(s(x''''), add(s(y''), add(0, add(0, x'0'''''')))) -> IFRM(eq(x'''', y''), s(x''''), add(s(y''), add(0, add(0, x'0''''''))))
RM(s(x''''), add(s(y''), add(0, add(s(y'''''''''), x'0'''''')))) -> IFRM(eq(x'''', y''), s(x''''), add(s(y''), add(0, add(s(y'''''''''), x'0''''''))))

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
Nar
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 13
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, 0, add(0, x'')) -> RM(0, x'')
RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))
IFRM(false, 0, add(s(x''''), x'')) -> RM(0, x'')
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(true, 0, add(0, x'')) -> RM(0, x'')
two new Dependency Pairs are created:

IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 15
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(false, 0, add(s(x''''), x'')) -> RM(0, x'')
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

RM(0, add(0, x)) -> IFRM(true, 0, add(0, x))
two new Dependency Pairs are created:

RM(0, add(0, add(0, x'''''))) -> IFRM(true, 0, add(0, add(0, x''''')))
RM(0, add(0, add(s(x''''''), x'0''))) -> IFRM(true, 0, add(0, add(s(x''''''), x'0'')))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 17
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
RM(0, add(0, add(s(x''''''), x'0''))) -> IFRM(true, 0, add(0, add(s(x''''''), x'0'')))
IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
RM(0, add(0, add(0, x'''''))) -> IFRM(true, 0, add(0, add(0, x''''')))
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
IFRM(false, 0, add(s(x''''), x'')) -> RM(0, x'')


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

IFRM(false, 0, add(s(x''''), x'')) -> RM(0, x'')
three new Dependency Pairs are created:

IFRM(false, 0, add(s(x''''), add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
IFRM(false, 0, add(s(x''''), add(0, add(0, x''''''')))) -> RM(0, add(0, add(0, x''''''')))
IFRM(false, 0, add(s(x''''), add(0, add(s(x''''''''), x'0'''')))) -> RM(0, add(0, add(s(x''''''''), x'0'''')))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 19
Forward Instantiation Transformation
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

IFRM(false, 0, add(s(x''''), add(0, add(s(x''''''''), x'0'''')))) -> RM(0, add(0, add(s(x''''''''), x'0'''')))
RM(0, add(0, add(s(x''''''), x'0''))) -> IFRM(true, 0, add(0, add(s(x''''''), x'0'')))
IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
RM(0, add(0, add(0, x'''''))) -> IFRM(true, 0, add(0, add(0, x''''')))
IFRM(false, 0, add(s(x''''), add(0, add(0, x''''''')))) -> RM(0, add(0, add(0, x''''''')))
IFRM(false, 0, add(s(x''''), add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




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

RM(0, add(s(x''), x)) -> IFRM(false, 0, add(s(x''), x))
three new Dependency Pairs are created:

RM(0, add(s(x'''), add(s(x'''''''), x'0''))) -> IFRM(false, 0, add(s(x'''), add(s(x'''''''), x'0'')))
RM(0, add(s(x'''), add(0, add(0, x''''''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(0, x'''''''''))))
RM(0, add(s(x'''), add(0, add(s(x''''''''''), x'0'''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(s(x''''''''''), x'0''''''))))

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
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 21
Argument Filtering and Ordering
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

RM(0, add(s(x'''), add(0, add(s(x''''''''''), x'0'''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(s(x''''''''''), x'0''''''))))
IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
RM(0, add(0, add(0, x'''''))) -> IFRM(true, 0, add(0, add(0, x''''')))
IFRM(false, 0, add(s(x''''), add(0, add(0, x''''''')))) -> RM(0, add(0, add(0, x''''''')))
RM(0, add(s(x'''), add(0, add(0, x''''''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(0, x'''''''''))))
IFRM(false, 0, add(s(x''''), add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
RM(0, add(s(x'''), add(s(x'''''''), x'0''))) -> IFRM(false, 0, add(s(x'''), add(s(x'''''''), x'0'')))
IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
RM(0, add(0, add(s(x''''''), x'0''))) -> IFRM(true, 0, add(0, add(s(x''''''), x'0'')))
IFRM(false, 0, add(s(x''''), add(0, add(s(x''''''''), x'0'''')))) -> RM(0, add(0, add(s(x''''''''), x'0'''')))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

IFRM(true, 0, add(0, add(0, x'''))) -> RM(0, add(0, x'''))
IFRM(false, 0, add(s(x''''), add(0, add(0, x''''''')))) -> RM(0, add(0, add(0, x''''''')))
IFRM(false, 0, add(s(x''''), add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
IFRM(true, 0, add(0, add(s(x''''), x'0))) -> RM(0, add(s(x''''), x'0))
IFRM(false, 0, add(s(x''''), add(0, add(s(x''''''''), x'0'''')))) -> RM(0, add(0, add(s(x''''''''), x'0'''')))


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:
RM(x1, x2) -> x2
add(x1, x2) -> add(x1, x2)
IFRM(x1, x2, x3) -> x3
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
           →DP Problem 10
Inst
             ...
               →DP Problem 23
Dependency Graph
       →DP Problem 5
Remaining
       →DP Problem 6
Remaining


Dependency Pairs:

RM(0, add(s(x'''), add(0, add(s(x''''''''''), x'0'''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(s(x''''''''''), x'0''''''))))
RM(0, add(0, add(0, x'''''))) -> IFRM(true, 0, add(0, add(0, x''''')))
RM(0, add(s(x'''), add(0, add(0, x''''''''')))) -> IFRM(false, 0, add(s(x'''), add(0, add(0, x'''''''''))))
RM(0, add(s(x'''), add(s(x'''''''), x'0''))) -> IFRM(false, 0, add(s(x'''), add(s(x'''''''), x'0'')))
RM(0, add(0, add(s(x''''''), x'0''))) -> IFRM(true, 0, add(0, add(s(x''''''), x'0'')))


Rules:


eq(0, 0) -> true
eq(0, s(x)) -> false
eq(s(x), 0) -> false
eq(s(x), s(y)) -> eq(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
min(add(n, nil)) -> n
min(add(n, add(m, x))) -> ifmin(le(n, m), add(n, add(m, x)))
ifmin(true, add(n, add(m, x))) -> min(add(n, x))
ifmin(false, add(n, add(m, x))) -> min(add(m, x))
rm(n, nil) -> nil
rm(n, add(m, x)) -> ifrm(eq(n, m), n, add(m, x))
ifrm(true, n, add(m, x)) -> rm(n, x)
ifrm(false, n, add(m, x)) -> add(m, rm(n, x))
minsort(nil, nil) -> nil
minsort(add(n, x), y) -> ifminsort(eq(n, min(add(n, x))), add(n, x), y)
ifminsort(true, add(n, x), y) -> add(n, minsort(app(rm(n, x), y), nil))
ifminsort(false, add(n, x), y) -> minsort(x, add(n, y))


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
Nar
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Nar
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:08 minutes