Termination Proof using AProVE

Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

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

one new Dependency Pair is created:

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

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(s(s(X''''))), s(s(s(Y'''')))) -> LE(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(LE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

two new Dependency Pairs are created:

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), 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

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

three new Dependency Pairs are created:

MINUS(s(0), s(Y''')) -> IFMINUS(true, s(0), s(Y'''))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), 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

             ↳Nar

...

               →DP Problem 8

                 ↳Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

IFMINUS(false, s(X), Y) -> MINUS(X, Y)

three new Dependency Pairs are created:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))
IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))

The transformation is resulting in three new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))

one new Dependency Pair is created:

IFMINUS(false, s(s(s(X''''))), 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

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 12

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(s(s(X''''))), s(s(Y''''))) -> MINUS(s(s(X'''')), s(s(Y'''')))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(s(X''''))), s(s(Y''''))) -> MINUS(s(s(X'''')), s(s(Y'''')))

Additionally, the following usable rules for innermost can be oriented:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(IFMINUS(x₁, x₂, x₃)) = x₂

POL(false) = 0

POL(MINUS(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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 7

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))

one new Dependency Pair is created:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(IFMINUS(x₁, x₂, x₃)) = x₂

POL(false) = 0

POL(MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 17

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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 7

             ↳Nar

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)

one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pairs:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)

one new Dependency Pair is created:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 15

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)
IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(IFMINUS(x₁, x₂, x₃)) = x₂

POL(false) = 0

POL(MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 18

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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(minus(X, Y), s(Y))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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(minus(X, Y), s(Y))

two new Dependency Pairs are created:

QUOT(s(0), s(Y'')) -> QUOT(0, s(Y''))
QUOT(s(s(X'')), s(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Narrowing Transformation

Dependency Pair:

QUOT(s(s(X'')), s(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), s(Y''))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(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(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), s(Y''))

two new Dependency Pairs are created:

QUOT(s(s(X''')), s(0)) -> QUOT(ifMinus(false, s(X'''), 0), s(0))
QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 20

                 ↳Rewriting Transformation

Dependency Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

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

one new Dependency Pair is created:

QUOT(s(s(X''')), s(0)) -> QUOT(s(minus(X''', 0)), 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 19

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

Dependency Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost can be oriented:

minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = 0

POL(ifMinus(x₁, x₂, x₃)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 21

                 ↳Polynomial Ordering

Dependency Pair:

QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))

Additionally, the following usable rules for innermost can be oriented:

ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁

POL(0) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = 0

POL(ifMinus(x₁, x₂, x₃)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

Dependency Pair:

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(0)	= 0
POL(IFMINUS(x₁, x₂, x₃))	= x₂
POL(false)	= 0
POL(MINUS(x₁, x₂))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le(x₁, x₂))	= 0

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(false)	= 0
POL(minus(x₁, x₂))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le(x₁, x₂))	= 0
POL(ifMinus(x₁, x₂, x₃))	= x₂