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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))
F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))


Rules:


f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y





The following dependency pairs can be strictly oriented:

F(f(x, y, z), u, f(x, y, v)) -> F(z, u, v)
F(f(x, y, z), u, f(x, y, v)) -> F(x, y, f(z, u, v))


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))=  0  
  POL(f(x1, x2, x3))=  1 + x1 + x3  
  POL(F(x1, x2, x3))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(f(x, y, z), u, f(x, y, v)) -> f(x, y, f(z, u, v))
f(x, y, y) -> y
f(x, y, g(y)) -> x
f(x, x, y) -> x
f(g(x), x, y) -> y





Using the Dependency Graph resulted in no new DP problems.

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