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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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) -> IF_MINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) -> LE(s(x), y)
IF_MINUS(false, s(x), y) -> MINUS(x, y)
MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
IF_MOD(true, 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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 w.r.t. to the implicit AFS 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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:

IF_MINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x), y) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

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

two new Dependency Pairs are created:

MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF_MINUS(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'')) -> IF_MINUS(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF_MINUS(false, s(x''), 0)
IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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'')) -> IF_MINUS(le(x'', y''), s(x''), s(y''))

three new Dependency Pairs are created:

MINUS(s(0), s(y''')) -> IF_MINUS(true, s(0), s(y'''))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(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'))) -> IF_MINUS(le(x', y'), s(s(x')), s(s(y')))
MINUS(s(s(x')), s(0)) -> IF_MINUS(false, s(s(x')), s(0))
IF_MINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x''), 0) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

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

three new Dependency Pairs are created:

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)
IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), 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:

IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))

one new Dependency Pair is created:

IF_MINUS(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:

IF_MINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(s(s(x''')), s(s(y'''')))
MINUS(s(s(x')), s(s(y'))) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = 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

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'))) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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:

IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
MINUS(s(s(x')), s(0)) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))

one new Dependency Pair is created:

IF_MINUS(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:

IF_MINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))
MINUS(s(s(x')), s(0)) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₂

POL(0) = 0

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)) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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:

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)
MINUS(s(x''), 0) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MINUS(false, s(x'), 0) -> MINUS(x', 0)

one new Dependency Pair is created:

IF_MINUS(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:

IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)
MINUS(s(x''), 0) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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) -> IF_MINUS(false, s(x''), 0)

one new Dependency Pair is created:

MINUS(s(s(x'''''')), 0) -> IF_MINUS(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) -> IF_MINUS(false, s(s(x'''''')), 0)
IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(IF_MINUS(x₁, x₂, x₃)) = x₂

POL(0) = 0

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) -> IF_MINUS(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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 Pairs:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IF_MOD(le(y, x), 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))

three new Dependency Pairs are created:

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
MOD(s(0), s(s(x''))) -> IF_MOD(false, s(0), s(s(x'')))
MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

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 Pairs:

MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))
MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
IF_MOD(true, s(x), s(y)) -> MOD(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

two new Dependency Pairs are created:

IF_MOD(true, s(0), s(y'')) -> MOD(0, s(y''))
IF_MOD(true, s(s(x'')), s(y'')) -> MOD(if_minus(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

             ↳Nar

...

               →DP Problem 20

                 ↳Narrowing Transformation

Dependency Pairs:

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))
IF_MOD(true, s(s(x'')), s(y'')) -> MOD(if_minus(le(s(x''), y''), s(x''), y''), s(y''))
MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

MOD(s(s(y'')), s(s(x''))) -> IF_MOD(le(x'', y''), s(s(y'')), s(s(x'')))

three new Dependency Pairs are created:

MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
MOD(s(s(0)), s(s(s(x')))) -> IF_MOD(false, s(s(0)), s(s(s(x'))))
MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

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 21

                 ↳Narrowing Transformation

Dependency Pairs:

MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))
MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
IF_MOD(true, s(s(x'')), s(y'')) -> MOD(if_minus(le(s(x''), y''), s(x''), y''), s(y''))
MOD(s(x'), s(0)) -> IF_MOD(true, 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MOD(true, s(s(x'')), s(y'')) -> MOD(if_minus(le(s(x''), y''), s(x''), y''), s(y''))

two new Dependency Pairs are created:

IF_MOD(true, s(s(x''')), s(0)) -> MOD(if_minus(false, s(x'''), 0), s(0))
IF_MOD(true, s(s(x''')), s(s(y'))) -> MOD(if_minus(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 22

                 ↳Polynomial Ordering

Dependency Pairs:

MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
IF_MOD(true, s(s(x''')), s(s(y'))) -> MOD(if_minus(le(x''', y'), s(x'''), s(y')), s(s(y')))
MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MOD(true, s(s(x''')), s(s(y'))) -> MOD(if_minus(le(x''', y'), s(x'''), s(y')), s(s(y')))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(MOD(x₁, x₂)) = x₁

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(IF_MOD(x₁, x₂, x₃)) = x₂

POL(if_minus(x₁, x₂, x₃)) = x₂

POL(le(x₁, x₂)) = 0

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 26

                 ↳Dependency Graph

Dependency Pairs:

MOD(s(s(y''')), s(s(0))) -> IF_MOD(true, s(s(y''')), s(s(0)))
MOD(s(s(s(y'))), s(s(s(x')))) -> IF_MOD(le(x', y'), s(s(s(y'))), s(s(s(x'))))

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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

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 19

             ↳Nar

...

               →DP Problem 23

                 ↳Rewriting Transformation

Dependency Pairs:

IF_MOD(true, s(s(x''')), s(0)) -> MOD(if_minus(false, s(x'''), 0), s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

IF_MOD(true, s(s(x''')), s(0)) -> MOD(if_minus(false, s(x'''), 0), s(0))

one new Dependency Pair is created:

IF_MOD(true, s(s(x''')), s(0)) -> MOD(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 24

                 ↳Forward Instantiation Transformation

Dependency Pairs:

IF_MOD(true, s(s(x''')), s(0)) -> MOD(s(minus(x''', 0)), s(0))
MOD(s(x'), s(0)) -> IF_MOD(true, 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

MOD(s(x'), s(0)) -> IF_MOD(true, s(x'), s(0))

one new Dependency Pair is created:

MOD(s(s(x''''')), s(0)) -> IF_MOD(true, 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 3

         ↳Nar

           →DP Problem 19

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

Dependency Pairs:

MOD(s(s(x''''')), s(0)) -> IF_MOD(true, s(s(x''''')), s(0))
IF_MOD(true, s(s(x''')), s(0)) -> MOD(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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_MOD(true, s(s(x''')), s(0)) -> MOD(s(minus(x''', 0)), s(0))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

minus(0, y) -> 0
minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(MOD(x₁, x₂)) = x₁

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(IF_MOD(x₁, x₂, x₃)) = x₂

POL(if_minus(x₁, x₂, x₃)) = x₂

POL(le(x₁, x₂)) = 0

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 27

                 ↳Dependency Graph

Dependency Pair:

MOD(s(s(x''''')), s(0)) -> IF_MOD(true, 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) -> if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) -> 0
if_minus(false, s(x), y) -> s(minus(x, y))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(IF_MINUS(x₁, x₂, x₃))	= 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(0)	= 0
POL(false)	= 0
POL(MOD(x₁, x₂))	= x₁
POL(minus(x₁, x₂))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(IF_MOD(x₁, x₂, x₃))	= x₂
POL(if_minus(x₁, x₂, x₃))	= x₂
POL(le(x₁, x₂))	= 0