Term Rewriting System R:
[x, y, z, l]
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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

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)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(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)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
SUM(nil) -> 0'(#)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules using the Ce-refinement 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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(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(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pairs:

+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules using the Ce-refinement 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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(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(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 6
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pairs:

+'(+(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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules using the Ce-refinement 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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(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(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 7
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

+'(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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))





The following dependency pair can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 8
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:

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