Termination Proof using AProVE

Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
IF_QUOT(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(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

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

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 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

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 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

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 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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

Strategy:

innermost

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

QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

two new Dependency Pairs are created:

QUOT(0, s(y')) -> IF_QUOT(false, 0, s(y'))
QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pairs:

QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)

Rules:

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

Strategy:

innermost

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

IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

IF_QUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))

Rules:

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

Strategy:

innermost

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

QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))

three new Dependency Pairs are created:

QUOT(s(y'''), s(0)) -> IF_QUOT(true, s(y'''), s(0))
QUOT(s(0), s(s(x'))) -> IF_QUOT(false, s(0), s(s(x')))
QUOT(s(s(y')), s(s(x'))) -> IF_QUOT(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

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

Dependency Pairs:

IF_QUOT(true, s(s(x')), s(s(y'))) -> QUOT(minus(x', y'), s(s(y')))
QUOT(s(s(y')), s(s(x'))) -> IF_QUOT(le(x', y'), s(s(y')), s(s(x')))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

minus(x, 0) -> x
minus(s(x), s(y)) -> 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)) = x₁

POL(0) = 0

POL(false) = 0

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(IF_QUOT(x₁, x₂, x₃)) = x₂

POL(le(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 15

                 ↳Dependency Graph

Dependency Pair:

QUOT(s(s(y')), s(s(x'))) -> IF_QUOT(le(x', y'), s(s(y')), s(s(x')))

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 14

                 ↳Polynomial Ordering

Dependency Pairs:

IF_QUOT(true, s(x'''), s(0)) -> QUOT(x''', s(0))
QUOT(s(y'''), s(0)) -> IF_QUOT(true, s(y'''), s(0))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF_QUOT(true, s(x'''), s(0)) -> QUOT(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)) = x₁

POL(0) = 0

POL(true) = 0

POL(IF_QUOT(x₁, x₂, x₃)) = x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

QUOT(s(y'''), s(0)) -> IF_QUOT(true, s(y'''), s(0))

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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

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