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: Polynomial ordering with Polynomial interpretation:

POL(PLUS(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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: Polynomial ordering with Polynomial interpretation:

POL(MIN(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(s(x₁)) = 1 + x₁

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: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(s(x₁)) = 1 + x₁

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

                 ↳Argument Filtering and Ordering

           →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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(plus(x₁, x₂)) = x₁ + x₂

POL(0) = 1

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 11

             ↳FwdInst

           →DP Problem 12

             ↳Nar

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

           →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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(Z) = 0

POL(plus(x₁, x₂)) = x₁ + x₂

POL(0) = 1

POL(min(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳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

                 ↳Argument Filtering and Ordering

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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(Z) = 0

POL(plus(x₁, x₂)) = x₁ + x₂

POL(0) = 1

POL(min(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

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

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(PLUS(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(MIN(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(Z)	= 0
POL(plus(x₁, x₂))	= x₁ + x₂
POL(0)	= 1
POL(min(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁