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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(g(x), a) -> F(x, g(a))
F(g(x), g(y)) -> H(g(y), x, g(y))
H(g(x), y, z) -> F(y, h(x, y, z))
H(g(x), y, z) -> H(x, y, z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

H(g(x), y, z) -> H(x, y, z)
H(g(x), y, z) -> F(y, h(x, y, z))
F(g(x), g(y)) -> H(g(y), x, g(y))
F(g(x), a) -> F(x, g(a))


Rules:


f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z





The following dependency pairs can be strictly oriented:

H(g(x), y, z) -> F(y, h(x, y, z))
F(g(x), a) -> F(x, g(a))


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(g(x1))=  1 + x1  
  POL(h(x1, x2, x3))=  0  
  POL(a)=  0  
  POL(H(x1, x2, x3))=  1 + x2  
  POL(f(x1, x2))=  0  
  POL(F(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

H(g(x), y, z) -> H(x, y, z)
F(g(x), g(y)) -> H(g(y), x, g(y))


Rules:


f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z





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


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


Dependency Pair:

H(g(x), y, z) -> H(x, y, z)


Rules:


f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z





The following dependency pair can be strictly oriented:

H(g(x), y, z) -> 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(g(x1))=  1 + x1  
  POL(H(x1, x2, x3))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z





Using the Dependency Graph resulted in no new DP problems.

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