Termination Proof using AProVE

Term Rewriting System R:
[x, y, z, k, l]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(minus(x, y), z) -> PLUS(y, z)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
PLUS(s(x), y) -> PLUS(x, y)
APP(cons(x, l), k) -> APP(l, k)
SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) -> PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) -> APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) -> SUM(cons(x, cons(y, k)))

Furthermore, R contains six SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

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

one new Dependency Pair is created:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')

one new Dependency Pair is created:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(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:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

APP(cons(x, l), k) -> APP(l, k)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

APP(cons(x, l), k) -> APP(l, k)

one new Dependency Pair is created:

APP(cons(x, cons(x'', l'')), k'') -> APP(cons(x'', l''), k'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

APP(cons(x, cons(x'', l'')), k'') -> APP(cons(x'', l''), k'')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

APP(cons(x, cons(x'', l'')), k'') -> APP(cons(x'', l''), k'')

one new Dependency Pair is created:

APP(cons(x, cons(x'''', cons(x''''', l''''))), k'''') -> APP(cons(x'''', cons(x''''', l'''')), k'''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

APP(cons(x, cons(x'''', cons(x''''', l''''))), k'''') -> APP(cons(x'''', cons(x''''', l'''')), k'''')

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(cons(x, cons(x'''', cons(x''''', l''''))), k'''') -> APP(cons(x'''', cons(x''''', l'''')), k'''')

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)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

one new Dependency Pair is created:

MINUS(minus(x, s(x'')), z') -> MINUS(x, s(plus(x'', z')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 13

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

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

one new Dependency Pair is created:

MINUS(s(s(x'')), s(s(y''))) -> MINUS(s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

MINUS(s(s(x'')), s(s(y''))) -> MINUS(s(x''), s(y''))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

MINUS(s(s(x'')), s(s(y''))) -> MINUS(s(x''), s(y''))

one new Dependency Pair is created:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(s(s(x'''')), s(s(y'''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(s(s(x'''')), s(s(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:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

The following usable rules for innermost can be oriented:

plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

SUM(x₁) -> SUM(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
plus(x₁, x₂) -> x₂
s(x₁) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

           →DP Problem 17

             ↳Dependency Graph

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

three new Dependency Pairs are created:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(s(x'')), s(s(y''))) -> QUOT(minus(x'', y''), s(s(y'')))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(x'')), s(s(y''))) -> QUOT(minus(x'', y''), s(s(y'')))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))
QUOT(s(x''), s(0)) -> QUOT(x'', s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(x'')), s(s(y''))) -> QUOT(minus(x'', y''), s(s(y'')))

three new Dependency Pairs are created:

QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 19

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))

one new Dependency Pair is created:

QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))

two new Dependency Pairs are created:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 21

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))

three new Dependency Pairs are created:

QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 22

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))

one new Dependency Pair is created:

QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 23

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))

two new Dependency Pairs are created:

QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 24

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))

three new Dependency Pairs are created:

QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 25

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))

one new Dependency Pair is created:

QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 26

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))

two new Dependency Pairs are created:

QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 27

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 28

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))

two new Dependency Pairs are created:

QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 29

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))

three new Dependency Pairs are created:

QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 30

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))

one new Dependency Pair is created:

QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 31

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))

two new Dependency Pairs are created:

QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 32

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 33

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))

two new Dependency Pairs are created:

QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 34

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 35

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))

two new Dependency Pairs are created:

QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 36

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 37

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 38

                 ↳Narrowing Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 39

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))

three new Dependency Pairs are created:

QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 40

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

three new Dependency Pairs are created:

QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 41

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))

four new Dependency Pairs are created:

QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 42

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))

five new Dependency Pairs are created:

QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 43

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))

four new Dependency Pairs are created:

QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 44

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))

four new Dependency Pairs are created:

QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 45

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))

five new Dependency Pairs are created:

QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 46

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))

five new Dependency Pairs are created:

QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 47

                 ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:

innermost

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

QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))

five new Dependency Pairs are created:

QUOT(s(s(s(s(minus(x''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(s(x''''''''''))))), s(0)) -> QUOT(s(s(s(s(x'''''''''')))), s(0))
QUOT(s(s(s(s(s(minus(x'''''''''', s(x'0'''''''))))))), s(0)) -> QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''''')))))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(s(s(s(x0''''''))))))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(s(s(s(x0'''''')))))))), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
QUOT(s(s(s(s(minus(x'''''''''', s(s(s(s(x0''''''))))))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(s(s(s(x0'''''')))))))), s(0))
QUOT(s(s(s(s(s(minus(x'''''''''', s(x'0'''''''))))))), s(0)) -> QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''''')))))), s(0))
QUOT(s(s(s(s(s(x''''''''''))))), s(0)) -> QUOT(s(s(s(s(x'''''''''')))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:
innermost
Dependency Pair:
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
QUOT(s(s(s(s(minus(x'''''''''', s(s(s(s(x0''''''))))))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(s(s(s(x0'''''')))))))), s(0))
QUOT(s(s(s(s(s(minus(x'''''''''', s(x'0'''''''))))))), s(0)) -> QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''''')))))), s(0))
QUOT(s(s(s(s(s(x''''''''''))))), s(0)) -> QUOT(s(s(s(s(x'''''''''')))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:
innermost
Dependency Pair:
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:40 minutes