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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(x, c(x), c(y)) -> F(y, y, f(y, x, y))
F(x, c(x), c(y)) -> F(y, x, y)
F(s(x), y, z) -> F(x, s(c(y)), c(z))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

F(s(x), y, z) -> F(x, s(c(y)), c(z))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y





The following dependency pair can be strictly oriented:

F(s(x), y, z) -> F(x, s(c(y)), c(z))


Additionally, the following rules can be oriented:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  0  
  POL(g(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  
  POL(f(x1, x2, x3))=  0  
  POL(F(x1, x2, x3))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y





Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pairs:

F(x, c(x), c(y)) -> F(y, x, y)
F(x, c(x), c(y)) -> F(y, y, f(y, x, y))


Rules:


f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y




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