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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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 w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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 w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 8
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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 w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
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
AFS
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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
Argument Filtering and Ordering
       →DP Problem 5
AFS
       →DP Problem 6
Nar


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




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
IFRM(x1, x2, x3) -> x3
add(x1, x2) -> add(x1, x2)
RM(x1, x2) -> x2


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


Dependency Pair:

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




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
Argument Filtering and Ordering
       →DP Problem 6
Nar


Dependency Pairs:

IFMIN(false, add(n, add(m, x))) -> MIN(add(m, x))
IFMIN(true, add(n, add(m, x))) -> MIN(add(n, x))
MIN(add(n, add(m, x))) -> IFMIN(le(n, m), add(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




The following dependency pairs can be strictly oriented:

IFMIN(false, add(n, add(m, x))) -> MIN(add(m, x))
IFMIN(true, add(n, add(m, x))) -> MIN(add(n, x))


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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
AFS
           →DP Problem 11
Dependency Graph
       →DP Problem 6
Nar


Dependency Pair:

MIN(add(n, add(m, x))) -> IFMIN(le(n, m), add(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




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
AFS
       →DP Problem 6
Narrowing Transformation


Dependency Pairs:

IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
IFMINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) -> IFMINSORT(eq(n, min(add(n, x))), add(n, 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




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

MINSORT(add(n, x), y) -> IFMINSORT(eq(n, min(add(n, x))), add(n, x), y)
two new Dependency Pairs are created:

MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Narrowing Transformation


Dependency Pairs:

MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), y)
IFMINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, 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




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

IFMINSORT(true, add(n, x), y) -> MINSORT(app(rm(n, x), y), nil)
two new Dependency Pairs are created:

IFMINSORT(true, add(n'', nil), y) -> MINSORT(app(nil, y), nil)
IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 13
Rewriting Transformation


Dependency Pairs:

IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)
IFMINSORT(true, add(n'', nil), y) -> MINSORT(app(nil, y), nil)
MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', 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




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

IFMINSORT(true, add(n'', nil), y) -> MINSORT(app(nil, y), nil)
one new Dependency Pair is created:

IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 14
Instantiation Transformation


Dependency Pairs:

IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)
MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), y)
IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)


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

IFMINSORT(false, add(n, x), y) -> MINSORT(x, add(n, y))
two new Dependency Pairs are created:

IFMINSORT(false, add(n', nil), y'') -> MINSORT(nil, add(n', y''))
IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 15
Instantiation Transformation


Dependency Pairs:

IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', y''))
IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)
MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), y)
MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)


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

MINSORT(add(n'', nil), y) -> IFMINSORT(eq(n'', n''), add(n'', nil), y)
two new Dependency Pairs are created:

MINSORT(add(n''', nil), nil) -> IFMINSORT(eq(n''', n'''), add(n''', nil), nil)
MINSORT(add(n'''', nil), add(n''''', y'''')) -> IFMINSORT(eq(n'''', n''''), add(n'''', nil), add(n''''', 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 16
Instantiation Transformation


Dependency Pairs:

MINSORT(add(n'''', nil), add(n''''', y'''')) -> IFMINSORT(eq(n'''', n''''), add(n'''', nil), add(n''''', y''''))
IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)
MINSORT(add(n''', nil), nil) -> IFMINSORT(eq(n''', n'''), add(n''', nil), nil)
IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)
MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), y)
IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', 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




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

MINSORT(add(n'', add(m', x'')), y) -> IFMINSORT(eq(n'', ifmin(le(n'', m'), add(n'', add(m', x'')))), add(n'', add(m', x'')), y)
two new Dependency Pairs are created:

MINSORT(add(n''', add(m'', x''')), nil) -> IFMINSORT(eq(n''', ifmin(le(n''', m''), add(n''', add(m'', x''')))), add(n''', add(m'', x''')), nil)
MINSORT(add(n'''', add(m'', x''')), add(n''''', y'''')) -> IFMINSORT(eq(n'''', ifmin(le(n'''', m''), add(n'''', add(m'', x''')))), add(n'''', add(m'', x''')), add(n''''', 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 17
Instantiation Transformation


Dependency Pairs:

MINSORT(add(n'''', add(m'', x''')), add(n''''', y'''')) -> IFMINSORT(eq(n'''', ifmin(le(n'''', m''), add(n'''', add(m'', x''')))), add(n'''', add(m'', x''')), add(n''''', y''''))
IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', y''))
IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)
MINSORT(add(n''', add(m'', x''')), nil) -> IFMINSORT(eq(n''', ifmin(le(n''', m''), add(n''', add(m'', x''')))), add(n''', add(m'', x''')), nil)
MINSORT(add(n''', nil), nil) -> IFMINSORT(eq(n''', n'''), add(n''', nil), nil)
IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)
MINSORT(add(n'''', nil), add(n''''', y'''')) -> IFMINSORT(eq(n'''', n''''), add(n'''', nil), add(n''''', 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




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

IFMINSORT(true, add(n'', add(m', x'')), y) -> MINSORT(app(ifrm(eq(n'', m'), n'', add(m', x'')), y), nil)
two new Dependency Pairs are created:

IFMINSORT(true, add(n''', add(m'', x''')), nil) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), nil), nil)
IFMINSORT(true, add(n''', add(m'', x''')), add(n'''''''', y'''''')) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), add(n'''''''', y'''''')), 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 18
Instantiation Transformation


Dependency Pairs:

IFMINSORT(true, add(n''', add(m'', x''')), add(n'''''''', y'''''')) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), add(n'''''''', y'''''')), nil)
IFMINSORT(true, add(n''', add(m'', x''')), nil) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), nil), nil)
MINSORT(add(n''', add(m'', x''')), nil) -> IFMINSORT(eq(n''', ifmin(le(n''', m''), add(n''', add(m'', x''')))), add(n''', add(m'', x''')), nil)
MINSORT(add(n''', nil), nil) -> IFMINSORT(eq(n''', n'''), add(n''', nil), nil)
IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)
MINSORT(add(n'''', nil), add(n''''', y'''')) -> IFMINSORT(eq(n'''', n''''), add(n'''', nil), add(n''''', y''''))
IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', y''))
MINSORT(add(n'''', add(m'', x''')), add(n''''', y'''')) -> IFMINSORT(eq(n'''', ifmin(le(n'''', m''), add(n'''', add(m'', x''')))), add(n'''', add(m'', x''')), add(n''''', 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




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

IFMINSORT(true, add(n'', nil), y) -> MINSORT(y, nil)
two new Dependency Pairs are created:

IFMINSORT(true, add(n''', nil), nil) -> MINSORT(nil, nil)
IFMINSORT(true, add(n''', nil), add(n'''''''', y'''''')) -> MINSORT(add(n'''''''', y''''''), 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
AFS
       →DP Problem 6
Nar
           →DP Problem 12
Nar
             ...
               →DP Problem 19
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IFMINSORT(true, add(n''', add(m'', x''')), nil) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), nil), nil)
MINSORT(add(n'''', add(m'', x''')), add(n''''', y'''')) -> IFMINSORT(eq(n'''', ifmin(le(n'''', m''), add(n'''', add(m'', x''')))), add(n'''', add(m'', x''')), add(n''''', y''''))
IFMINSORT(true, add(n''', nil), add(n'''''''', y'''''')) -> MINSORT(add(n'''''''', y''''''), nil)
MINSORT(add(n'''', nil), add(n''''', y'''')) -> IFMINSORT(eq(n'''', n''''), add(n'''', nil), add(n''''', y''''))
IFMINSORT(false, add(n', add(m''', x''''')), y'') -> MINSORT(add(m''', x'''''), add(n', y''))
MINSORT(add(n''', add(m'', x''')), nil) -> IFMINSORT(eq(n''', ifmin(le(n''', m''), add(n''', add(m'', x''')))), add(n''', add(m'', x''')), nil)
IFMINSORT(true, add(n''', add(m'', x''')), add(n'''''''', y'''''')) -> MINSORT(app(ifrm(eq(n''', m''), n''', add(m'', x''')), add(n'''''''', y'''''')), nil)


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



Innermost Termination of R could not be shown.
Duration:
0:10 minutes