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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
+'(s(x), y) -> +'(x, y)
*'(x, s(y)) -> +'(x, *(x, y))
*'(x, s(y)) -> *'(x, y)
F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
F(s(x)) -> -'(*(s(s(0)), s(x)), s(s(x)))
F(s(x)) -> *'(s(s(0)), s(x))

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(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:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(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:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 6
Argument Filtering and Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(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 w.r.t. to the AFS 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)


   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Forward Instantiation Transformation
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(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), y) -> +'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(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'')), y'') -> +'(s(x''), y'')
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 9
Argument Filtering and Ordering
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw


Dependency Pair:


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Forward Instantiation Transformation
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

*'(x, s(y)) -> *'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
Forward Instantiation Transformation
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

*'(x'', s(s(y''))) -> *'(x'', s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
FwdInst
             ...
               →DP Problem 12
Argument Filtering and Ordering
       →DP Problem 4
Rw


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS 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)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
           →DP Problem 11
FwdInst
             ...
               →DP Problem 13
Dependency Graph
       →DP Problem 4
Rw


Dependency Pair:


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 15
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 16
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 17
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 18
Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




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

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))
one new Dependency Pair is created:

F(s(x)) -> F(-(*(s(s(0)), x), x))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
FwdInst
       →DP Problem 4
Rw
           →DP Problem 14
Rw
             ...
               →DP Problem 19
Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), x), x))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes