Termination Proof using AProVE

Term Rewriting System R:
[y, x]
leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LEQ(s(x), s(y)) -> LEQ(x, y)
-'(s(x), s(y)) -> -'(x, y)
MOD(s(x), s(y)) -> IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) -> LEQ(y, x)
MOD(s(x), s(y)) -> MOD(-(s(x), s(y)), s(y))
MOD(s(x), s(y)) -> -'(s(x), s(y))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), 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

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

one new Dependency Pair is created:

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

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), 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

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

one new Dependency Pair is created:

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

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LEQ(s(s(s(x''''))), s(s(s(y'''')))) -> LEQ(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(LEQ(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

LEQ(x₁, x₂) -> LEQ(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

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

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), 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

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

one new Dependency Pair is created:

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

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), 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

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

one new Dependency Pair is created:

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

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(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(-'(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

-'(x₁, x₂) -> -'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rewriting Transformation

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

         ↳Rw

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), 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)) -> MOD(-(x, y), s(y))

two new Dependency Pairs are created:

MOD(s(x''), s(0)) -> MOD(x'', s(0))
MOD(s(s(x'')), s(s(y''))) -> MOD(-(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

         ↳Rw

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MOD(s(x''), s(0)) -> MOD(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(MOD(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

MOD(x₁, x₂) -> MOD(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Dependency Graph

Dependency Pair:

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

Dependency Pair:

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

Rules:

leq(0, y) -> true
leq(s(x), 0) -> false
leq(s(x), s(y)) -> leq(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MOD(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

MOD(x₁, x₂) -> MOD(x₁, x₂)
s(x₁) -> s(x₁)
-(x₁, x₂) -> x₁

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

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

POL(-'(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

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