Termination Proof using AProVE

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

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

no new Dependency Pairs are created.
The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*(x₁, x₂)) = x₂

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

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

Strategy:

innermost

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + x₁ + x₂

POL(*(x₁, x₂)) = x₂

POL(+(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

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

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