Term Rewriting System R:
[x, y, z]
+(0, 0) -> 0
+(0, 1) -> 1
+(0, 2) -> 2
+(0, 3) -> 3
+(0, 4) -> 4
+(0, 5) -> 5
+(0, 6) -> 6
+(0, 7) -> 7
+(0, 8) -> 8
+(0, 9) -> 9
+(1, 0) -> 1
+(1, 1) -> 2
+(1, 2) -> 3
+(1, 3) -> 4
+(1, 4) -> 5
+(1, 5) -> 6
+(1, 6) -> 7
+(1, 7) -> 8
+(1, 8) -> 9
+(1, 9) -> c(1, 0)
+(2, 0) -> 2
+(2, 1) -> 3
+(2, 2) -> 4
+(2, 3) -> 5
+(2, 4) -> 6
+(2, 5) -> 7
+(2, 6) -> 8
+(2, 7) -> 9
+(2, 8) -> c(1, 0)
+(2, 9) -> c(1, 1)
+(3, 0) -> 3
+(3, 1) -> 4
+(3, 2) -> 5
+(3, 3) -> 6
+(3, 4) -> 7
+(3, 5) -> 8
+(3, 6) -> 9
+(3, 7) -> c(1, 0)
+(3, 8) -> c(1, 1)
+(3, 9) -> c(1, 2)
+(4, 0) -> 4
+(4, 1) -> 5
+(4, 2) -> 6
+(4, 3) -> 7
+(4, 4) -> 8
+(4, 5) -> 9
+(4, 6) -> c(1, 0)
+(4, 7) -> c(1, 1)
+(4, 8) -> c(1, 2)
+(4, 9) -> c(1, 3)
+(5, 0) -> 5
+(5, 1) -> 6
+(5, 2) -> 7
+(5, 3) -> 8
+(5, 4) -> 9
+(5, 5) -> c(1, 0)
+(5, 6) -> c(1, 1)
+(5, 7) -> c(1, 2)
+(5, 8) -> c(1, 3)
+(5, 9) -> c(1, 4)
+(6, 0) -> 6
+(6, 1) -> 7
+(6, 2) -> 8
+(6, 3) -> 9
+(6, 4) -> c(1, 0)
+(6, 5) -> c(1, 1)
+(6, 6) -> c(1, 2)
+(6, 7) -> c(1, 3)
+(6, 8) -> c(1, 4)
+(6, 9) -> c(1, 5)
+(7, 0) -> 7
+(7, 1) -> 8
+(7, 2) -> 9
+(7, 3) -> c(1, 0)
+(7, 4) -> c(1, 1)
+(7, 5) -> c(1, 2)
+(7, 6) -> c(1, 3)
+(7, 7) -> c(1, 4)
+(7, 8) -> c(1, 5)
+(7, 9) -> c(1, 6)
+(8, 0) -> 8
+(8, 1) -> 9
+(8, 2) -> c(1, 0)
+(8, 3) -> c(1, 1)
+(8, 4) -> c(1, 2)
+(8, 5) -> c(1, 3)
+(8, 6) -> c(1, 4)
+(8, 7) -> c(1, 5)
+(8, 8) -> c(1, 6)
+(8, 9) -> c(1, 7)
+(9, 0) -> 9
+(9, 1) -> c(1, 0)
+(9, 2) -> c(1, 1)
+(9, 3) -> c(1, 2)
+(9, 4) -> c(1, 3)
+(9, 5) -> c(1, 4)
+(9, 6) -> c(1, 5)
+(9, 7) -> c(1, 6)
+(9, 8) -> c(1, 7)
+(9, 9) -> c(1, 8)
+(x, c(y, z)) -> c(y, +(x, z))
+(c(x, y), z) -> c(x, +(y, z))
c(0, x) -> x
c(x, c(y, z)) -> c(+(x, y), z)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(1, 9) -> C(1, 0)
+'(2, 8) -> C(1, 0)
+'(2, 9) -> C(1, 1)
+'(3, 7) -> C(1, 0)
+'(3, 8) -> C(1, 1)
+'(3, 9) -> C(1, 2)
+'(4, 6) -> C(1, 0)
+'(4, 7) -> C(1, 1)
+'(4, 8) -> C(1, 2)
+'(4, 9) -> C(1, 3)
+'(5, 5) -> C(1, 0)
+'(5, 6) -> C(1, 1)
+'(5, 7) -> C(1, 2)
+'(5, 8) -> C(1, 3)
+'(5, 9) -> C(1, 4)
+'(6, 4) -> C(1, 0)
+'(6, 5) -> C(1, 1)
+'(6, 6) -> C(1, 2)
+'(6, 7) -> C(1, 3)
+'(6, 8) -> C(1, 4)
+'(6, 9) -> C(1, 5)
+'(7, 3) -> C(1, 0)
+'(7, 4) -> C(1, 1)
+'(7, 5) -> C(1, 2)
+'(7, 6) -> C(1, 3)
+'(7, 7) -> C(1, 4)
+'(7, 8) -> C(1, 5)
+'(7, 9) -> C(1, 6)
+'(8, 2) -> C(1, 0)
+'(8, 3) -> C(1, 1)
+'(8, 4) -> C(1, 2)
+'(8, 5) -> C(1, 3)
+'(8, 6) -> C(1, 4)
+'(8, 7) -> C(1, 5)
+'(8, 8) -> C(1, 6)
+'(8, 9) -> C(1, 7)
+'(9, 1) -> C(1, 0)
+'(9, 2) -> C(1, 1)
+'(9, 3) -> C(1, 2)
+'(9, 4) -> C(1, 3)
+'(9, 5) -> C(1, 4)
+'(9, 6) -> C(1, 5)
+'(9, 7) -> C(1, 6)
+'(9, 8) -> C(1, 7)
+'(9, 9) -> C(1, 8)
+'(x, c(y, z)) -> C(y, +(x, z))
+'(x, c(y, z)) -> +'(x, z)
+'(c(x, y), z) -> C(x, +(y, z))
+'(c(x, y), z) -> +'(y, z)
C(x, c(y, z)) -> C(+(x, y), z)
C(x, c(y, z)) -> +'(x, y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 0) -> 0
+(0, 1) -> 1
+(0, 2) -> 2
+(0, 3) -> 3
+(0, 4) -> 4
+(0, 5) -> 5
+(0, 6) -> 6
+(0, 7) -> 7
+(0, 8) -> 8
+(0, 9) -> 9
+(1, 0) -> 1
+(1, 1) -> 2
+(1, 2) -> 3
+(1, 3) -> 4
+(1, 4) -> 5
+(1, 5) -> 6
+(1, 6) -> 7
+(1, 7) -> 8
+(1, 8) -> 9
+(1, 9) -> c(1, 0)
+(2, 0) -> 2
+(2, 1) -> 3
+(2, 2) -> 4
+(2, 3) -> 5
+(2, 4) -> 6
+(2, 5) -> 7
+(2, 6) -> 8
+(2, 7) -> 9
+(2, 8) -> c(1, 0)
+(2, 9) -> c(1, 1)
+(3, 0) -> 3
+(3, 1) -> 4
+(3, 2) -> 5
+(3, 3) -> 6
+(3, 4) -> 7
+(3, 5) -> 8
+(3, 6) -> 9
+(3, 7) -> c(1, 0)
+(3, 8) -> c(1, 1)
+(3, 9) -> c(1, 2)
+(4, 0) -> 4
+(4, 1) -> 5
+(4, 2) -> 6
+(4, 3) -> 7
+(4, 4) -> 8
+(4, 5) -> 9
+(4, 6) -> c(1, 0)
+(4, 7) -> c(1, 1)
+(4, 8) -> c(1, 2)
+(4, 9) -> c(1, 3)
+(5, 0) -> 5
+(5, 1) -> 6
+(5, 2) -> 7
+(5, 3) -> 8
+(5, 4) -> 9
+(5, 5) -> c(1, 0)
+(5, 6) -> c(1, 1)
+(5, 7) -> c(1, 2)
+(5, 8) -> c(1, 3)
+(5, 9) -> c(1, 4)
+(6, 0) -> 6
+(6, 1) -> 7
+(6, 2) -> 8
+(6, 3) -> 9
+(6, 4) -> c(1, 0)
+(6, 5) -> c(1, 1)
+(6, 6) -> c(1, 2)
+(6, 7) -> c(1, 3)
+(6, 8) -> c(1, 4)
+(6, 9) -> c(1, 5)
+(7, 0) -> 7
+(7, 1) -> 8
+(7, 2) -> 9
+(7, 3) -> c(1, 0)
+(7, 4) -> c(1, 1)
+(7, 5) -> c(1, 2)
+(7, 6) -> c(1, 3)
+(7, 7) -> c(1, 4)
+(7, 8) -> c(1, 5)
+(7, 9) -> c(1, 6)
+(8, 0) -> 8
+(8, 1) -> 9
+(8, 2) -> c(1, 0)
+(8, 3) -> c(1, 1)
+(8, 4) -> c(1, 2)
+(8, 5) -> c(1, 3)
+(8, 6) -> c(1, 4)
+(8, 7) -> c(1, 5)
+(8, 8) -> c(1, 6)
+(8, 9) -> c(1, 7)
+(9, 0) -> 9
+(9, 1) -> c(1, 0)
+(9, 2) -> c(1, 1)
+(9, 3) -> c(1, 2)
+(9, 4) -> c(1, 3)
+(9, 5) -> c(1, 4)
+(9, 6) -> c(1, 5)
+(9, 7) -> c(1, 6)
+(9, 8) -> c(1, 7)
+(9, 9) -> c(1, 8)
+(x, c(y, z)) -> c(y, +(x, z))
+(c(x, y), z) -> c(x, +(y, z))
c(0, x) -> x
c(x, c(y, z)) -> c(+(x, y), z)


Strategy:

innermost




We have the following set of usable rules:

+(4, 4) -> 8
+(5, 2) -> 7
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(1, 6) -> 7
+(2, 4) -> 6
+(6, 4) -> c(1, 0)
+(9, 9) -> c(1, 8)
+(6, 5) -> c(1, 1)
+(0, 3) -> 3
+(2, 9) -> c(1, 1)
+(3, 1) -> 4
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(9, 0) -> 9
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(7, 5) -> c(1, 2)
+(0, 2) -> 2
+(7, 4) -> c(1, 1)
+(2, 5) -> 7
+(4, 6) -> c(1, 0)
+(0, 0) -> 0
+(5, 9) -> c(1, 4)
+(8, 1) -> 9
+(0, 7) -> 7
+(9, 4) -> c(1, 3)
+(1, 4) -> 5
+(3, 4) -> 7
+(7, 6) -> c(1, 3)
+(4, 5) -> 9
+(4, 9) -> c(1, 3)
+(3, 5) -> 8
+(5, 3) -> 8
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(5, 8) -> c(1, 3)
+(8, 5) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(1, 7) -> 8
+(2, 8) -> c(1, 0)
+(3, 3) -> 6
+(8, 0) -> 8
+(c(x, y), z) -> c(x, +(y, z))
+(2, 2) -> 4
+(4, 8) -> c(1, 2)
+(5, 5) -> c(1, 0)
+(0, 8) -> 8
+(3, 0) -> 3
+(9, 1) -> c(1, 0)
+(4, 2) -> 6
+(8, 2) -> c(1, 0)
+(8, 6) -> c(1, 4)
+(6, 8) -> c(1, 4)
+(9, 2) -> c(1, 1)
+(7, 0) -> 7
+(1, 8) -> 9
+(1, 1) -> 2
+(6, 0) -> 6
+(4, 3) -> 7
+(9, 6) -> c(1, 5)
c(x, c(y, z)) -> c(+(x, y), z)
+(6, 9) -> c(1, 5)
+(9, 5) -> c(1, 4)
+(9, 8) -> c(1, 7)
+(2, 0) -> 2
+(5, 0) -> 5
+(1, 0) -> 1
+(3, 7) -> c(1, 0)
+(7, 1) -> 8
+(2, 6) -> 8
+(9, 7) -> c(1, 6)
+(6, 1) -> 7
+(6, 3) -> 9
+(7, 9) -> c(1, 6)
+(5, 4) -> 9
+(1, 2) -> 3
+(9, 3) -> c(1, 2)
+(0, 6) -> 6
+(0, 9) -> 9
+(8, 7) -> c(1, 5)
+(2, 1) -> 3
+(7, 3) -> c(1, 0)
+(0, 1) -> 1
+(4, 1) -> 5
+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(8, 9) -> c(1, 7)
+(1, 5) -> 6
c(0, x) -> x
+(7, 7) -> c(1, 4)
+(6, 2) -> 8
+(4, 7) -> c(1, 1)
+(1, 3) -> 4
+(2, 7) -> 9
+(8, 4) -> c(1, 2)
+(7, 2) -> 9
+(1, 9) -> c(1, 0)
+(5, 7) -> c(1, 2)
+(5, 1) -> 6
+(3, 6) -> 9
+(3, 2) -> 5
+(2, 3) -> 5
+(3, 9) -> c(1, 2)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  1  
  POL(8)=  1  
  POL(5)=  1  
  POL(2)=  1  
  POL(6)=  1  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  1  
  POL(9)=  1  
  POL(3)=  1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  1  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(4, 4) -> 8
+(5, 2) -> 7
+(1, 6) -> 7
+(2, 4) -> 6
+(6, 4) -> c(1, 0)
+(3, 1) -> 4
+(2, 5) -> 7
+(4, 6) -> c(1, 0)
+(8, 1) -> 9
+(1, 4) -> 5
+(3, 4) -> 7
+(4, 5) -> 9
+(3, 5) -> 8
+(5, 3) -> 8
+(1, 7) -> 8
+(2, 8) -> c(1, 0)
+(3, 3) -> 6
+(2, 2) -> 4
+(5, 5) -> c(1, 0)
+(9, 1) -> c(1, 0)
+(4, 2) -> 6
+(8, 2) -> c(1, 0)
+(1, 8) -> 9
+(1, 1) -> 2
+(4, 3) -> 7
+(3, 7) -> c(1, 0)
+(7, 1) -> 8
+(2, 6) -> 8
+(6, 1) -> 7
+(6, 3) -> 9
+(5, 4) -> 9
+(1, 2) -> 3
+(2, 1) -> 3
+(7, 3) -> c(1, 0)
+(4, 1) -> 5
+(1, 5) -> 6
+(6, 2) -> 8
+(1, 3) -> 4
+(2, 7) -> 9
+(7, 2) -> 9
+(1, 9) -> c(1, 0)
+(5, 1) -> 6
+(3, 6) -> 9
+(3, 2) -> 5
+(2, 3) -> 5


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(9, 9) -> c(1, 8)
+(6, 5) -> c(1, 1)
+(0, 3) -> 3
+(2, 9) -> c(1, 1)
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(9, 0) -> 9
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(7, 5) -> c(1, 2)
+(0, 2) -> 2
+(7, 4) -> c(1, 1)
+(0, 0) -> 0
+(5, 9) -> c(1, 4)
+(0, 7) -> 7
+(9, 4) -> c(1, 3)
+(7, 6) -> c(1, 3)
+(4, 9) -> c(1, 3)
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(5, 8) -> c(1, 3)
+(8, 5) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(8, 0) -> 8
+(c(x, y), z) -> c(x, +(y, z))
+(4, 8) -> c(1, 2)
+(0, 8) -> 8
+(3, 0) -> 3
+(8, 6) -> c(1, 4)
+(6, 8) -> c(1, 4)
+(9, 2) -> c(1, 1)
+(7, 0) -> 7
+(6, 0) -> 6
+(9, 6) -> c(1, 5)
+(6, 9) -> c(1, 5)
+(9, 5) -> c(1, 4)
+(9, 8) -> c(1, 7)
+(2, 0) -> 2
+(5, 0) -> 5
+(1, 0) -> 1
+(9, 7) -> c(1, 6)
+(7, 9) -> c(1, 6)
+(9, 3) -> c(1, 2)
+(0, 6) -> 6
+(0, 9) -> 9
+(8, 7) -> c(1, 5)
+(0, 1) -> 1
+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(8, 9) -> c(1, 7)
+(7, 7) -> c(1, 4)
+(4, 7) -> c(1, 1)
+(8, 4) -> c(1, 2)
+(5, 7) -> c(1, 2)
+(3, 9) -> c(1, 2)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(9, 2) -> c(1, 1)
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(9, 6) -> c(1, 5)
c(x, c(y, z)) -> c(+(x, y), z)
+(9, 9) -> c(1, 8)
+(6, 9) -> c(1, 5)
+(9, 5) -> c(1, 4)
+(9, 8) -> c(1, 7)
+(6, 5) -> c(1, 1)
+(2, 0) -> 2
+(0, 3) -> 3
+(5, 0) -> 5
+(1, 0) -> 1
+(2, 9) -> c(1, 1)
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(9, 7) -> c(1, 6)
+(9, 0) -> 9
+(7, 9) -> c(1, 6)
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(9, 3) -> c(1, 2)
+(0, 6) -> 6
+(0, 2) -> 2
+(7, 5) -> c(1, 2)
+(7, 4) -> c(1, 1)
+(0, 9) -> 9
+(0, 0) -> 0
+(8, 7) -> c(1, 5)
+(5, 9) -> c(1, 4)
+(9, 4) -> c(1, 3)
+(0, 7) -> 7
+(0, 1) -> 1
+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(7, 6) -> c(1, 3)
+(8, 9) -> c(1, 7)
+(4, 9) -> c(1, 3)
c(0, x) -> x
+(7, 7) -> c(1, 4)
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(4, 7) -> c(1, 1)
+(5, 8) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(8, 5) -> c(1, 3)
+(8, 4) -> c(1, 2)
+(8, 0) -> 8
+(c(x, y), z) -> c(x, +(y, z))
+(5, 7) -> c(1, 2)
+(4, 8) -> c(1, 2)
+(0, 8) -> 8
+(3, 0) -> 3
+(8, 6) -> c(1, 4)
+(3, 9) -> c(1, 2)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(9)=  1  
  POL(3)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(9, 2) -> c(1, 1)
+(9, 6) -> c(1, 5)
+(9, 9) -> c(1, 8)
+(6, 9) -> c(1, 5)
+(9, 5) -> c(1, 4)
+(9, 8) -> c(1, 7)
+(2, 9) -> c(1, 1)
+(9, 7) -> c(1, 6)
+(7, 9) -> c(1, 6)
+(9, 3) -> c(1, 2)
+(5, 9) -> c(1, 4)
+(9, 4) -> c(1, 3)
+(8, 9) -> c(1, 7)
+(4, 9) -> c(1, 3)
+(3, 9) -> c(1, 2)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 3
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(6, 5) -> c(1, 1)
+(2, 0) -> 2
+(0, 3) -> 3
+(5, 0) -> 5
+(1, 0) -> 1
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(9, 0) -> 9
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(0, 6) -> 6
+(0, 2) -> 2
+(7, 5) -> c(1, 2)
+(7, 4) -> c(1, 1)
+(0, 9) -> 9
+(0, 0) -> 0
+(8, 7) -> c(1, 5)
+(0, 7) -> 7
+(0, 1) -> 1
+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(7, 6) -> c(1, 3)
+(7, 7) -> c(1, 4)
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(4, 7) -> c(1, 1)
+(5, 8) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(8, 5) -> c(1, 3)
+(8, 4) -> c(1, 2)
+(8, 0) -> 8
+(c(x, y), z) -> c(x, +(y, z))
+(5, 7) -> c(1, 2)
+(4, 8) -> c(1, 2)
+(0, 8) -> 8
+(3, 0) -> 3
+(8, 6) -> c(1, 4)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(7, 6) -> c(1, 3)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x
+(7, 7) -> c(1, 4)
+(6, 5) -> c(1, 1)
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(4, 7) -> c(1, 1)
+(0, 5) -> 5
+(5, 8) -> c(1, 3)
+(8, 5) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(5, 7) -> c(1, 2)
+(0, 2) -> 2
+(7, 5) -> c(1, 2)
+(4, 8) -> c(1, 2)
+(7, 4) -> c(1, 1)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 7) -> c(1, 5)
+(8, 6) -> c(1, 4)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(9)=  0  
  POL(3)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  1  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(7, 8) -> c(1, 5)
+(6, 7) -> c(1, 3)
+(7, 6) -> c(1, 3)
+(7, 7) -> c(1, 4)
+(4, 7) -> c(1, 1)
+(5, 7) -> c(1, 2)
+(7, 5) -> c(1, 2)
+(7, 4) -> c(1, 1)
+(8, 7) -> c(1, 5)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 4
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(6, 5) -> c(1, 1)
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(0, 5) -> 5
+(5, 8) -> c(1, 3)
+(8, 5) -> c(1, 3)
+(8, 8) -> c(1, 6)
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x
+(6, 5) -> c(1, 1)
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(0, 5) -> 5
+(5, 8) -> c(1, 3)
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 8) -> c(1, 6)
+(8, 5) -> c(1, 3)
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(5, 6) -> c(1, 1)
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  1  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(9)=  0  
  POL(3)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(6, 5) -> c(1, 1)
+(5, 8) -> c(1, 3)
+(8, 5) -> c(1, 3)
+(5, 6) -> c(1, 1)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 8) -> c(1, 6)
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 3) -> c(1, 1)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(3, 8) -> c(1, 1)
+(0, 5) -> 5
+(8, 8) -> c(1, 6)
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(9)=  0  
  POL(3)=  1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(8, 3) -> c(1, 1)
+(3, 8) -> c(1, 1)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(0, 5) -> 5
+(8, 8) -> c(1, 6)
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 7) -> 7
+(6, 8) -> c(1, 4)
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(8, 8) -> c(1, 6)
+(8, 4) -> c(1, 2)
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(4, 8) -> c(1, 2)
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
+(8, 6) -> c(1, 4)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  1  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(9)=  0  
  POL(3)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 3, 9, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(6, 8) -> c(1, 4)
+(8, 8) -> c(1, 6)
+(8, 4) -> c(1, 2)
+(4, 8) -> c(1, 2)
+(8, 6) -> c(1, 4)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 7) -> 7
+(7, 0) -> 7
+(0, 1) -> 1
+(6, 0) -> 6
+(0, 4) -> 4
+(2, 0) -> 2
+(5, 0) -> 5
+(0, 3) -> 3
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(9, 0) -> 9
+(8, 0) -> 8
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(0, 0) -> 0
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 5) -> 5
+(0, 7) -> 7
+(x, c(y, z)) -> c(y, +(x, z))
+(0, 1) -> 1
+(7, 0) -> 7
+(9, 0) -> 9
+(8, 0) -> 8
+(0, 4) -> 4
+(6, 0) -> 6
+(4, 0) -> 4
c(x, c(y, z)) -> c(+(x, y), z)
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
c(0, x) -> x
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(2, 0) -> 2
+(0, 3) -> 3
+(5, 0) -> 5
+(0, 0) -> 0
+(1, 0) -> 1
+(6, 6) -> c(1, 2)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  1  
  POL(c(x1, x2))=  x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(3)=  0  
  POL(9)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 9, 3, +, +', 7}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(6, 6) -> c(1, 2)


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 8
                 ↳Modular Removal of Rules


Dependency Pairs:

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


Rules:


+(0, 5) -> 5
+(0, 7) -> 7
+(x, c(y, z)) -> c(y, +(x, z))
+(0, 1) -> 1
+(7, 0) -> 7
+(9, 0) -> 9
+(8, 0) -> 8
+(0, 4) -> 4
+(6, 0) -> 6
+(4, 0) -> 4
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(2, 0) -> 2
+(0, 3) -> 3
+(5, 0) -> 5
+(0, 0) -> 0
+(1, 0) -> 1
c(x, c(y, z)) -> c(+(x, y), z)
c(0, x) -> x


Strategy:

innermost




We have the following set of usable rules:

+(0, 5) -> 5
+(x, c(y, z)) -> c(y, +(x, z))
+(0, 7) -> 7
+(7, 0) -> 7
+(0, 1) -> 1
+(9, 0) -> 9
+(6, 0) -> 6
+(0, 4) -> 4
+(8, 0) -> 8
+(4, 0) -> 4
c(x, c(y, z)) -> c(+(x, y), z)
+(c(x, y), z) -> c(x, +(y, z))
+(0, 6) -> 6
+(0, 2) -> 2
c(0, x) -> x
+(0, 9) -> 9
+(0, 8) -> 8
+(3, 0) -> 3
+(2, 0) -> 2
+(0, 3) -> 3
+(5, 0) -> 5
+(0, 0) -> 0
+(1, 0) -> 1
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(4)=  0  
  POL(8)=  0  
  POL(5)=  0  
  POL(2)=  0  
  POL(6)=  0  
  POL(c(x1, x2))=  1 + x1 + x2  
  POL(C(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(1)=  0  
  POL(3)=  0  
  POL(9)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  
  POL(7)=  0  

We have the following set D of usable symbols: {4, 8, 5, 2, 6, c, 0, C, 1, 9, 3, +, +', 7}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


Innermost Termination of R successfully shown.
Duration:
0:21 minutes