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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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), s(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, x, y) -> MOD(minus(x, y), y)
IF_MOD(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), y)
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, 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)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, 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(MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳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, x, y) -> MOD(minus(x, y), y)

Rules:

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Instantiation Transformation

Dependency Pairs:

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''))
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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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_MOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))

two new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Polynomial Ordering

Dependency Pairs:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

POL(MOD(x₁, x₂)) = x₁

POL(true) = 0

POL(IF_MOD(x₁, x₂, x₃)) = x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pairs:

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

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

minus(x, 0) -> x
minus(s(x), s(y)) -> 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(le(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Dependency Graph

Dependency Pair:

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(x, 0) -> x
minus(s(x), s(y)) -> 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, x, y) -> mod(minus(x, y), 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:
0:00 minutes

POL(0)	= 0
POL(MOD(x₁, x₂))	= x₁
POL(true)	= 0
POL(IF_MOD(x₁, x₂, x₃))	= x₂
POL(s(x₁))	= 1 + x₁

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(le(x₁, x₂))	= 0