Term Rewriting System R:
[x, y, z]
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(x, s(y)) -> +'(x, y)
+'(s(x), y) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
F(g(f(x))) -> F(h(s(0), x))
F(g(h(x, y))) -> F(h(s(x), y))
F(h(x, h(y, z))) -> F(h(+(x, y), z))
F(h(x, h(y, z))) -> +'(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





The following dependency pairs can be strictly oriented:

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


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(0)=  0  
  POL(s(x1))=  x1  
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem. 



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


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





The following dependency pair can be strictly oriented:

+'(s(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(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

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


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





The following dependency pair can be strictly oriented:

+'(x, s(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(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





The following dependency pair can be strictly oriented:

F(h(x, h(y, z))) -> F(h(+(x, y), z))


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(0)=  0  
  POL(s(x1))=  0  
  POL(h(x1, x2))=  1 + x2  
  POL(+(x1, x2))=  0  
  POL(F(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph


Dependency Pair:


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))





Using the Dependency Graph resulted in no new DP problems.

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