Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(*, x), app(app(+, y), z)) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z))
APP(app(*, x), app(app(+, y), z)) -> APP(+, app(app(*, x), y))
APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y)
APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z)
APP(app(*, app(app(+, y), z)), x) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z))
APP(app(*, app(app(+, y), z)), x) -> APP(+, app(app(*, x), y))
APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y)
APP(app(*, app(app(+, y), z)), x) -> APP(*, x)
APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z)
APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z))
APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z)
APP(app(*, app(app(*, x), y)), z) -> APP(*, y)
APP(app(+, app(app(+, x), y)), z) -> APP(app(+, x), app(app(+, y), z))
APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z)
APP(app(+, app(app(+, x), y)), z) -> APP(+, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳ATrans

Dependency Pair:

APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z)

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 4 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳A-Transformation

       →DP Problem 2

         ↳ATrans

Dependency Pair:

APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z)

Rule:

none

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳ATrans

...

               →DP Problem 4

                 ↳Size-Change Principle

       →DP Problem 2

         ↳ATrans

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	=	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	=	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

+(x₁, x₂) -> +(x₁, x₂)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳A-Transformation

Dependency Pairs:

APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z)
APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z))
APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z)
APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z)
APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y)
APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y)

Rules:

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

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Nar

...

               →DP Problem 6

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁ + x₁·x₂ + x₂

POL(*(x₁, x₂)) = x₁ + x₁·x₂ + x₂

POL(+(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Nar

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Nar

...

               →DP Problem 8

                 ↳Usable Rules (Innermost)

Dependency Pair:

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

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 4 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Nar

...

               →DP Problem 9

                 ↳Size-Change Principle

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	=	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	=	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

*(x₁, x₂) -> *(x₁, x₂)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:15 minutes

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