Term Rewriting System R:
[x, y]
O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(O(x), O(y)) -> O'(+(x, y))
+'(O(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> O'(+(+(x, y), I(0)))
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), I(y)) -> +'(x, y)
*'(O(x), y) -> O'(*(x, y))
*'(O(x), y) -> *'(x, y)
*'(I(x), y) -> +'(O(*(x, y)), y)
*'(I(x), y) -> O'(*(x, y))
*'(I(x), y) -> *'(x, y)
-'(O(x), O(y)) -> O'(-(x, y))
-'(O(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(I(x), I(y)) -> O'(-(x, y))
-'(I(x), I(y)) -> -'(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





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

+'(I(x), I(y)) -> +'(+(x, y), I(0))
six new Dependency Pairs are created:

+'(I(0), I(y')) -> +'(y', I(0))
+'(I(x''), I(0)) -> +'(x'', I(0))
+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'', y'')), I(0))
+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 4
Polynomial Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pairs:

+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))
+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'', y'')), I(0))
+'(I(x''), I(0)) -> +'(x'', I(0))
+'(I(0), I(y')) -> +'(y', I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pairs can be strictly oriented:

+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'', y'')), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(0)=  0  
  POL(O(x1))=  1 + x1  
  POL(+(x1, x2))=  0  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 4
Polo
             ...
               →DP Problem 5
Polynomial Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pairs:

+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))
+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(x''), I(0)) -> +'(x'', I(0))
+'(I(0), I(y')) -> +'(y', I(0))
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pairs can be strictly oriented:

+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))
+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'', y'')), I(0))
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  1 + x1  
  POL(0)=  0  
  POL(O(x1))=  0  
  POL(+(x1, x2))=  0  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 4
Polo
             ...
               →DP Problem 6
Instantiation Transformation
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pairs:

+'(I(x''), I(0)) -> +'(x'', I(0))
+'(I(0), I(y')) -> +'(y', I(0))


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





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

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

+'(I(0), I(0)) -> +'(0, I(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 4
Polo
             ...
               →DP Problem 7
Polynomial Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pair:

+'(I(x''), I(0)) -> +'(x'', I(0))


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pair can be strictly oriented:

+'(I(x''), I(0)) -> +'(x'', I(0))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  1 + x1  
  POL(0)=  0  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 4
Polo
             ...
               →DP Problem 8
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Polo


Dependency Pair:


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Narrowing Transformation
       →DP Problem 3
Polo


Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





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

-'(O(x), I(y)) -> -'(-(x, y), I(1))
six new Dependency Pairs are created:

-'(O(x''), I(0)) -> -'(x'', I(1))
-'(O(0), I(y')) -> -'(0, I(1))
-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'', y''), I(1))), I(1))
-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'', y'')), I(1))
-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 9
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pairs:

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'', y'')), I(1))
-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'', y''), I(1))), I(1))
-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(x''), I(0)) -> -'(x'', I(1))
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), O(y)) -> -'(x, y)
-'(I(x), I(y)) -> -'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pair can be strictly oriented:

-'(O(x''), I(0)) -> -'(x'', I(1))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(0)=  1  
  POL(-'(x1, x2))=  x2  
  POL(1)=  0  
  POL(O(x1))=  x1  
  POL(-(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 9
Polo
             ...
               →DP Problem 10
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pairs:

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'', y'')), I(1))
-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'', y''), I(1))), I(1))
-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'', y'')), I(1))
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), O(y)) -> -'(x, y)
-'(I(x), I(y)) -> -'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pairs can be strictly oriented:

-'(O(I(x'')), I(O(y''))) -> -'(I(-(x'', y'')), I(1))
-'(O(O(x'')), I(O(y''))) -> -'(O(-(x'', y'')), I(1))
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), O(y)) -> -'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  x1  
  POL(0)=  0  
  POL(-'(x1, x2))=  x2  
  POL(1)=  0  
  POL(O(x1))=  1 + x1  
  POL(-(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 9
Polo
             ...
               →DP Problem 11
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pairs:

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'', y''), I(1))), I(1))
-'(O(x), I(y)) -> -'(x, y)
-'(I(x), I(y)) -> -'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pairs can be strictly oriented:

-'(O(I(x'')), I(I(y''))) -> -'(O(-(x'', y'')), I(1))
-'(O(O(x'')), I(I(y''))) -> -'(I(-(-(x'', y''), I(1))), I(1))
-'(O(x), I(y)) -> -'(x, y)
-'(I(x), I(y)) -> -'(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1))=  1 + x1  
  POL(0)=  0  
  POL(-'(x1, x2))=  x2  
  POL(1)=  0  
  POL(O(x1))=  0  
  POL(-(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
           →DP Problem 9
Polo
             ...
               →DP Problem 12
Dependency Graph
       →DP Problem 3
Polo


Dependency Pair:


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
       →DP Problem 3
Polynomial Ordering


Dependency Pairs:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
       →DP Problem 3
Polo
           →DP Problem 13
Polynomial Ordering


Dependency Pair:

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


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Nar
       →DP Problem 3
Polo
           →DP Problem 13
Polo
             ...
               →DP Problem 14
Dependency Graph


Dependency Pair:


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)
-(x, 0) -> x
-(0, x) -> 0
-(O(x), O(y)) -> O(-(x, y))
-(O(x), I(y)) -> I(-(-(x, y), I(1)))
-(I(x), O(y)) -> I(-(x, y))
-(I(x), I(y)) -> O(-(x, y))





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes