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
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →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
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 6
Dependency Graph
       →DP Problem 3
Polo
       →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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(*'(x1, x2))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 7
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 9
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 10
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 11
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 12
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
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 13
Narrowing Transformation


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




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

F(s(x)) -> F(-(*(s(s(0)), x), x))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 14
Rewriting Transformation


Dependency Pairs:

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


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

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 15
Rewriting Transformation


Dependency Pairs:

F(s(0)) -> F(-(0, 0))
F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

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

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 16
Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 17
Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 18
Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 19
Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), 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 Rewriting SCC transformation can be performed.
As a result of transforming the rule

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Rw
           →DP Problem 8
Rw
             ...
               →DP Problem 20
Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

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



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