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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →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




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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS
       →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




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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS
       →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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and 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))
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(y)) -> *'(x, y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →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))
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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 9
                 ↳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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 10
                 ↳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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 11
                 ↳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
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 12
                 ↳Narrowing 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 Narrowing 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)))))
eight new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 13
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 20
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 21
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 22
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 23
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 24
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 25
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 26
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 27
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 28
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 29
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 30
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 31
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 32
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 33
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 34
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 35
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 36
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 37
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 38
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 39
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 40
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 41
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 42
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 43
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 44
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 45
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 46
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 47
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 48
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 49
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 50
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 51
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 52
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 53
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 54
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 55
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 56
                 ↳Rewriting Transformation


Dependency Pairs:

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 57
                 ↳Narrowing Transformation


Dependency Pairs:

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

F(s(0)) -> F(-(0, s(0)))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 58
                 ↳Narrowing Transformation


Dependency Pairs:

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


Strategy:

innermost




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

F(s(0)) -> F(-(0, s(0)))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 59
                 ↳Narrowing Transformation


Dependency Pairs:

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

F(s(0)) -> F(-(0, s(0)))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 60
                 ↳Narrowing Transformation


Dependency Pairs:

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


Strategy:

innermost




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

F(s(0)) -> F(-(0, s(0)))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Rw
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 61
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(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:
0:06 minutes