Termination Proof using AProVE

Term Rewriting System R:
[Y, X]
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

PLUS(s(X), Y) -> PLUS(X, Y)
MIN(s(X), s(Y)) -> MIN(X, Y)
MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(min(X, Y), Z) -> PLUS(Y, Z)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

PLUS(s(s(X'')), Y'') -> PLUS(s(X''), Y'')

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

PLUS(s(s(s(X''''))), Y'''') -> PLUS(s(s(X'''')), Y'''')

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))

one new Dependency Pair is created:

MIN(min(X, s(X'')), Z) -> MIN(X, s(plus(X'', Z)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

MIN(s(X), s(Y)) -> MIN(X, Y)

one new Dependency Pair is created:

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

one new Dependency Pair is created:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(X'''')), s(s(Y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(X'''')), s(s(Y'''')))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(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:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(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(min(X'', Y''), s(s(Y'')))
QUOT(s(min(X'', Y'')), s(Z)) -> QUOT(min(X'', plus(Y'', Z)), s(Z))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳Forward Instantiation Transformation

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 14

                 ↳Argument Filtering and Ordering

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 19

                 ↳Dependency Graph

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

Dependency Pair:

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(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(min(X', Y'), s(s(s(Y'))))
QUOT(s(s(min(X', Y'))), s(s(Z))) -> QUOT(min(X', plus(Y', Z)), s(s(Z)))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(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(min(X'', Y''), s(s(s(s(Y'')))))
QUOT(s(s(s(min(X'', Y'')))), s(s(s(Z)))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(Z))))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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))))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 26

                 ↳Forward Instantiation Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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))))

one new Dependency Pair is created:

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))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 33

                 ↳Argument Filtering and Ordering

           →DP Problem 13

             ↳Nar

Dependency Pair:

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))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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))))

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:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 48

                 ↳Dependency Graph

           →DP Problem 13

             ↳Nar

Dependency Pair:

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 22

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(s(X'')))), s(s(s(s(Y''))))) -> QUOT(min(X'', Y''), s(s(s(s(Y'')))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(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(min(X', Y'), s(s(s(s(s(Y'))))))
QUOT(s(s(s(s(min(X', Y'))))), s(s(s(s(Z))))) -> QUOT(min(X', plus(Y', Z)), s(s(s(s(Z)))))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 27

                 ↳Forward Instantiation Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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)))))

one new Dependency Pair is 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)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 34

                 ↳Argument Filtering and Ordering

           →DP Problem 13

             ↳Nar

Dependency Pair:

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)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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)))))

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:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 28

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(s(s(X'))))), s(s(s(s(s(Y')))))) -> QUOT(min(X', Y'), s(s(s(s(s(Y'))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(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(min(X'', Y''), s(s(s(s(s(s(Y'')))))))
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 35

                 ↳Argument Filtering and Ordering

           →DP Problem 13

             ↳Nar

Dependency Pair:

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

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 29

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(s(min(X', Y'))))), s(s(s(s(Z))))) -> QUOT(min(X', plus(Y', Z)), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

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(min(X', Y'))))), s(s(s(s(Z))))) -> QUOT(min(X', plus(Y', Z)), s(s(s(s(Z)))))

one new Dependency Pair is created:

QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 23

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(min(X'', Y'')))), s(s(s(Z)))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(s(min(X'', Y'')))), s(s(s(Z)))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(Z))))

one new Dependency Pair is created:

QUOT(s(s(s(min(X'', s(X'))))), s(s(s(Z)))) -> QUOT(min(X'', s(plus(X', Z))), s(s(s(Z))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 30

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(s(min(X'', s(X'))))), s(s(s(Z)))) -> QUOT(min(X'', s(plus(X', Z))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(s(min(X'', s(X'))))), s(s(s(Z)))) -> QUOT(min(X'', s(plus(X', Z))), s(s(s(Z))))

two new Dependency Pairs are created:

QUOT(s(s(s(min(X'', s(0))))), s(s(s(Z)))) -> QUOT(min(X'', s(Z)), s(s(s(Z))))
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 39

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pairs:

QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))
QUOT(s(s(s(min(X'', s(0))))), s(s(s(Z)))) -> QUOT(min(X'', s(Z)), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(s(min(X'', s(0))))), s(s(s(Z)))) -> QUOT(min(X'', s(Z)), s(s(s(Z))))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 17

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(min(X', Y'))), s(s(Z))) -> QUOT(min(X', plus(Y', Z)), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(min(X', Y'))), s(s(Z))) -> QUOT(min(X', plus(Y', Z)), s(s(Z)))

one new Dependency Pair is created:

QUOT(s(s(min(X', s(X'')))), s(s(Z))) -> QUOT(min(X', s(plus(X'', Z))), s(s(Z)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 24

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(min(X', s(X'')))), s(s(Z))) -> QUOT(min(X', s(plus(X'', Z))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(min(X', s(X'')))), s(s(Z))) -> QUOT(min(X', s(plus(X'', Z))), s(s(Z)))

two new Dependency Pairs are created:

QUOT(s(s(min(X', s(0)))), s(s(Z))) -> QUOT(min(X', s(Z)), s(s(Z)))
QUOT(s(s(min(X', s(s(X0))))), s(s(Z))) -> QUOT(min(X', s(s(plus(X0, Z)))), s(s(Z)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 31

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pairs:

QUOT(s(s(min(X', s(s(X0))))), s(s(Z))) -> QUOT(min(X', s(s(plus(X0, Z)))), s(s(Z)))
QUOT(s(s(min(X', s(0)))), s(s(Z))) -> QUOT(min(X', s(Z)), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(min(X', s(0)))), s(s(Z))) -> QUOT(min(X', s(Z)), s(s(Z)))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 40

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pair:

QUOT(s(s(min(X', s(s(X0))))), s(s(Z))) -> QUOT(min(X', s(s(plus(X0, Z)))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(min(X', s(s(X0))))), s(s(Z))) -> QUOT(min(X', s(s(plus(X0, Z)))), s(s(Z)))

two new Dependency Pairs are created:

QUOT(s(s(min(X', s(s(0))))), s(s(Z))) -> QUOT(min(X', s(s(Z))), s(s(Z)))
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 43

                 ↳Narrowing Transformation

           →DP Problem 13

             ↳Nar

Dependency Pairs:

QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))
QUOT(s(s(min(X', s(s(0))))), s(s(Z))) -> QUOT(min(X', s(s(Z))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(s(min(X', s(s(0))))), s(s(Z))) -> QUOT(min(X', s(s(Z))), s(s(Z)))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Narrowing Transformation

Dependency Pair:

QUOT(s(min(X'', Y'')), s(Z)) -> QUOT(min(X'', plus(Y'', Z)), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', Y'')), s(Z)) -> QUOT(min(X'', plus(Y'', Z)), s(Z))

one new Dependency Pair is created:

QUOT(s(min(X'', s(X'))), s(Z)) -> QUOT(min(X'', s(plus(X', Z))), s(Z))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 18

                 ↳Narrowing Transformation

Dependency Pair:

QUOT(s(min(X'', s(X'))), s(Z)) -> QUOT(min(X'', s(plus(X', Z))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(X'))), s(Z)) -> QUOT(min(X'', s(plus(X', Z))), s(Z))

two new Dependency Pairs are created:

QUOT(s(min(X'', s(0))), s(Z)) -> QUOT(min(X'', s(Z)), s(Z))
QUOT(s(min(X'', s(s(X0)))), s(Z)) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(Z))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 25

                 ↳Narrowing Transformation

Dependency Pairs:

QUOT(s(min(X'', s(s(X0)))), s(Z)) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(Z))
QUOT(s(min(X'', s(0))), s(Z)) -> QUOT(min(X'', s(Z)), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(0))), s(Z)) -> QUOT(min(X'', s(Z)), s(Z))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 32

                 ↳Narrowing Transformation

Dependency Pair:

QUOT(s(min(X'', s(s(X0)))), s(Z)) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(s(X0)))), s(Z)) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(Z))

two new Dependency Pairs are created:

QUOT(s(min(X'', s(s(0)))), s(Z)) -> QUOT(min(X'', s(s(Z))), s(Z))
QUOT(s(min(X'', s(s(s(X'))))), s(Z)) -> QUOT(min(X'', s(s(s(plus(X', Z))))), s(Z))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 41

                 ↳Narrowing Transformation

Dependency Pairs:

QUOT(s(min(X'', s(s(s(X'))))), s(Z)) -> QUOT(min(X'', s(s(s(plus(X', Z))))), s(Z))
QUOT(s(min(X'', s(s(0)))), s(Z)) -> QUOT(min(X'', s(s(Z))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(s(0)))), s(Z)) -> QUOT(min(X'', s(s(Z))), s(Z))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 44

                 ↳Narrowing Transformation

Dependency Pair:

QUOT(s(min(X'', s(s(s(X'))))), s(Z)) -> QUOT(min(X'', s(s(s(plus(X', Z))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(s(s(X'))))), s(Z)) -> QUOT(min(X'', s(s(s(plus(X', Z))))), s(Z))

two new Dependency Pairs are created:

QUOT(s(min(X'', s(s(s(0))))), s(Z)) -> QUOT(min(X'', s(s(s(Z)))), s(Z))
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 46

                 ↳Narrowing Transformation

Dependency Pairs:

QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))
QUOT(s(min(X'', s(s(s(0))))), s(Z)) -> QUOT(min(X'', s(s(s(Z)))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:

innermost

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

QUOT(s(min(X'', s(s(s(0))))), s(Z)) -> QUOT(min(X'', s(s(s(Z)))), s(Z))

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 47

                 ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(s(s(s(s(s(X'')))))), s(s(s(s(s(s(Y''))))))) -> QUOT(min(X'', Y''), s(s(s(s(s(s(Y'')))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(s(min(X'', Y'')))))), s(s(s(s(s(Z)))))) -> QUOT(min(X'', plus(Y'', Z)), s(s(s(s(s(Z))))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(s(min(X', s(X'')))))), s(s(s(s(Z))))) -> QUOT(min(X', s(plus(X'', Z))), s(s(s(s(Z)))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(s(min(X'', s(s(X0)))))), s(s(s(Z)))) -> QUOT(min(X'', s(s(plus(X0, Z)))), s(s(s(Z))))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(s(min(X', s(s(s(X'')))))), s(s(Z))) -> QUOT(min(X', s(s(s(plus(X'', Z))))), s(s(Z)))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost
Dependency Pair:
QUOT(s(min(X'', s(s(s(s(X0)))))), s(Z)) -> QUOT(min(X'', s(s(s(s(plus(X0, Z)))))), s(Z))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes