Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(+(x, y), z) -> +'(*(x, z), *(y, z))
*'(+(x, y), z) -> *'(x, z)
*'(+(x, y), z) -> *'(y, z)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

+'(+(x, y), z) -> +'(y, z)

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

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

+'(+(x, y), z) -> +'(y, z)

one new Dependency Pair is created:

+'(+(x, +(x'', y'')), z'') -> +'(+(x'', y''), z'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

+'(+(x, +(x'', y'')), z'') -> +'(+(x'', y''), z'')

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(+(x, +(x'', y'')), z'') -> +'(+(x'', y''), z'')

The following usable rules for innermost can be oriented:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(i(x₁)) = x₁

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

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

+'(x₁, x₂) -> +'(x₁, x₂)
+(x₁, x₂) -> +(x₁, x₂)
i(x₁) -> i(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳AFS

...

               →DP Problem 4

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pairs:

*'(+(x, y), z) -> *'(y, z)
*'(x, +(y, z)) -> *'(x, z)
*'(+(x, y), z) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

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

*'(x, +(y, z)) -> *'(x, y)

one new Dependency Pair is created:

*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

Dependency Pairs:

*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)
*'(+(x, y), z) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, z)
*'(+(x, y), z) -> *'(y, z)

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

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

*'(x, +(y, z)) -> *'(x, z)

three new Dependency Pairs are created:

*'(x'', +(y, +(y'', z''))) -> *'(x'', +(y'', z''))
*'(+(x'', y''), +(y, z'')) -> *'(+(x'', y''), z'')
*'(+(x'''', y''''), +(y, +(y0'', z''))) -> *'(+(x'''', y''''), +(y0'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

*'(+(x'''', y''''), +(y, +(y0'', z''))) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x'', y''), +(y, z'')) -> *'(+(x'', y''), z'')
*'(+(x, y), z) -> *'(y, z)
*'(x'', +(y, +(y'', z''))) -> *'(x'', +(y'', z''))
*'(+(x, y), z) -> *'(x, z)
*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

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

*'(+(x, y), z) -> *'(x, z)

one new Dependency Pair is created:

*'(+(x', y), +(y'', +(y'''', z''''))) -> *'(x', +(y'', +(y'''', z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

*'(+(x', y), +(y'', +(y'''', z''''))) -> *'(x', +(y'', +(y'''', z'''')))
*'(+(x'', y''), +(y, z'')) -> *'(+(x'', y''), z'')
*'(x'', +(y, +(y'', z''))) -> *'(x'', +(y'', z''))
*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)
*'(+(x, y), z) -> *'(y, z)
*'(+(x'''', y''''), +(y, +(y0'', z''))) -> *'(+(x'''', y''''), +(y0'', z''))

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

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

*'(+(x, y), z) -> *'(y, z)

six new Dependency Pairs are created:

*'(+(x, +(x'', y'')), z'') -> *'(+(x'', y''), z'')
*'(+(x, +(x'''', y'''')), +(y0'', z'')) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x, y0), +(y'', +(y'''', z''''))) -> *'(y0, +(y'', +(y'''', z'''')))
*'(+(x, +(x'''', y'''')), +(y'', z'''')) -> *'(+(x'''', y''''), +(y'', z''''))
*'(+(x, +(x'''''', y'''''')), +(y'', +(y0'''', z''''))) -> *'(+(x'''''', y''''''), +(y'', +(y0'''', z'''')))
*'(+(x, +(x''', y'')), +(y'''', +(y'''''', z''''''))) -> *'(+(x''', y''), +(y'''', +(y'''''', z'''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

Dependency Pairs:

*'(+(x, +(x''', y'')), +(y'''', +(y'''''', z''''''))) -> *'(+(x''', y''), +(y'''', +(y'''''', z'''''')))
*'(+(x, +(x'''''', y'''''')), +(y'', +(y0'''', z''''))) -> *'(+(x'''''', y''''''), +(y'', +(y0'''', z'''')))
*'(+(x, +(x'''', y'''')), +(y'', z'''')) -> *'(+(x'''', y''''), +(y'', z''''))
*'(+(x, y0), +(y'', +(y'''', z''''))) -> *'(y0, +(y'', +(y'''', z'''')))
*'(+(x, +(x'''', y'''')), +(y0'', z'')) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x'''', y''''), +(y, +(y0'', z''))) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x'', y''), +(y, z'')) -> *'(+(x'', y''), z'')
*'(+(x, +(x'', y'')), z'') -> *'(+(x'', y''), z'')
*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)
*'(x'', +(y, +(y'', z''))) -> *'(x'', +(y'', z''))
*'(+(x', y), +(y'', +(y'''', z''''))) -> *'(x', +(y'', +(y'''', z'''')))

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

*'(+(x, +(x''', y'')), +(y'''', +(y'''''', z''''''))) -> *'(+(x''', y''), +(y'''', +(y'''''', z'''''')))
*'(+(x, +(x'''''', y'''''')), +(y'', +(y0'''', z''''))) -> *'(+(x'''''', y''''''), +(y'', +(y0'''', z'''')))
*'(+(x, +(x'''', y'''')), +(y'', z'''')) -> *'(+(x'''', y''''), +(y'', z''''))
*'(+(x, y0), +(y'', +(y'''', z''''))) -> *'(y0, +(y'', +(y'''', z'''')))
*'(+(x, +(x'''', y'''')), +(y0'', z'')) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x'''', y''''), +(y, +(y0'', z''))) -> *'(+(x'''', y''''), +(y0'', z''))
*'(+(x'', y''), +(y, z'')) -> *'(+(x'', y''), z'')
*'(+(x, +(x'', y'')), z'') -> *'(+(x'', y''), z'')
*'(+(x'', y''), +(y0, z)) -> *'(+(x'', y''), y0)
*'(x'', +(y, +(y'', z''))) -> *'(x'', +(y'', z''))
*'(+(x', y), +(y'', +(y'''', z''''))) -> *'(x', +(y'', +(y'''', z'''')))

The following usable rules for innermost can be oriented:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(i(x₁)) = x₁

POL(*'(x₁, x₂)) = 1 + x₁ + x₂

POL(+(x₁, x₂)) = 1 + x₁ + x₂

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

*'(x₁, x₂) -> *'(x₁, x₂)
+(x₁, x₂) -> +(x₁, x₂)
i(x₁) -> i(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, i(x)) -> 0
+(+(x, y), z) -> +(x, +(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(+(x, y), z) -> +(*(x, z), *(y, z))

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(i(x₁))	= x₁
POL(+(x₁, x₂))	= 1 + x₁ + x₂
POL(+'(x₁, x₂))	= 1 + x₁ + x₂