Term Rewriting System R:
[x, y, z]
0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(j(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(j(x), j(y)) -> +'(x, y)
+'(1(x), j(y)) -> 0'(+(x, y))
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), 1(y)) -> 0'(+(x, y))
+'(j(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
OPP(0(x)) -> 0'(opp(x))
OPP(0(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(j(x)) -> OPP(x)
-'(x, y) -> +'(x, opp(y))
-'(x, y) -> OPP(y)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(j(x), y) -> -'(0(*(x, y)), y)
*'(j(x), y) -> 0'(*(x, y))
*'(j(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)

Furthermore, R contains three SCCs.


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


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(j(x), 1(y)) -> +'(x, y)
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(j(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

+'(j(x), 1(y)) -> +'(x, y)
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(j(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)


Additionally, the following usable rules for innermost can be oriented:

+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #


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

resulting in one new DP problem.



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


Dependency Pairs:

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


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules for innermost can be oriented:

+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Polo
             ...
               →DP Problem 5
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pairs:

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


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Polo
             ...
               →DP Problem 6
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pair:

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


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 4
Polo
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo


Dependency Pair:


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

OPP(j(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

OPP(j(x)) -> OPP(x)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0(x1))=  x1  
  POL(1(x1))=  x1  
  POL(OPP(x1))=  x1  
  POL(j(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 8
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pairs:

OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

OPP(1(x)) -> OPP(x)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0(x1))=  x1  
  POL(1(x1))=  1 + x1  
  POL(OPP(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 8
Polo
             ...
               →DP Problem 9
Polynomial Ordering
       →DP Problem 3
Polo


Dependency Pair:

OPP(0(x)) -> OPP(x)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

OPP(0(x)) -> OPP(x)


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0(x1))=  1 + x1  
  POL(OPP(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 8
Polo
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
Polo


Dependency Pair:


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



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


Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 11
Polo
             ...
               →DP Problem 12
Polynomial Ordering


Dependency Pairs:

*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 11
Polo
             ...
               →DP Problem 13
Polynomial Ordering


Dependency Pair:

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


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 11
Polo
             ...
               →DP Problem 14
Dependency Graph


Dependency Pair:


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes