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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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(-(p(*(s(x), s(x))), *(s(x), s(x))))
F(s(x)) -> -'(p(*(s(x), s(x))), *(s(x), s(x)))
F(s(x)) -> P(*(s(x), s(x)))
F(s(x)) -> *'(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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 that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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(-(p(*(s(x), s(x))), *(s(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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), 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(-(p(*(s(x), s(x))), *(s(x), s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(p(+(s(x), *(s(x), x))), *(s(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
             ↳Rewriting Transformation


Dependency Pair:

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

F(s(x)) -> F(-(p(s(+(x, *(s(x), x)))), *(s(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 15
                 ↳Rewriting Transformation


Dependency Pair:

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

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

F(s(x)) -> F(-(+(x, *(s(x), x)), +(s(x), *(s(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 17
                 ↳Rewriting Transformation


Dependency Pair:

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

F(s(x)) -> F(-(+(x, *(s(x), x)), s(+(x, *(s(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
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(s(x)) -> F(-(+(x, *(s(x), x)), s(+(x, *(s(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))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), s(x))))


Strategy:

innermost



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