Term Rewriting System R:
[x]
a(c(d(x))) -> c(x)
u(b(d(d(x)))) -> b(x)
v(a(a(x))) -> u(v(x))
v(a(c(x))) -> u(b(d(x)))
v(c(x)) -> b(x)
w(a(a(x))) -> u(w(x))
w(a(c(x))) -> u(b(d(x)))
w(c(x)) -> b(x)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

V(a(a(x))) -> U(v(x))
V(a(a(x))) -> V(x)
V(a(c(x))) -> U(b(d(x)))
W(a(a(x))) -> U(w(x))
W(a(a(x))) -> W(x)
W(a(c(x))) -> U(b(d(x)))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

V(a(a(x))) -> V(x)


Rules:


a(c(d(x))) -> c(x)
u(b(d(d(x)))) -> b(x)
v(a(a(x))) -> u(v(x))
v(a(c(x))) -> u(b(d(x)))
v(c(x)) -> b(x)
w(a(a(x))) -> u(w(x))
w(a(c(x))) -> u(b(d(x)))
w(c(x)) -> b(x)





The following dependency pair can be strictly oriented:

V(a(a(x))) -> V(x)


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(V(x1))=  1 + x1  
  POL(a(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


a(c(d(x))) -> c(x)
u(b(d(d(x)))) -> b(x)
v(a(a(x))) -> u(v(x))
v(a(c(x))) -> u(b(d(x)))
v(c(x)) -> b(x)
w(a(a(x))) -> u(w(x))
w(a(c(x))) -> u(b(d(x)))
w(c(x)) -> b(x)





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

W(a(a(x))) -> W(x)


Rules:


a(c(d(x))) -> c(x)
u(b(d(d(x)))) -> b(x)
v(a(a(x))) -> u(v(x))
v(a(c(x))) -> u(b(d(x)))
v(c(x)) -> b(x)
w(a(a(x))) -> u(w(x))
w(a(c(x))) -> u(b(d(x)))
w(c(x)) -> b(x)





The following dependency pair can be strictly oriented:

W(a(a(x))) -> W(x)


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(W(x1))=  1 + x1  
  POL(a(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


a(c(d(x))) -> c(x)
u(b(d(d(x)))) -> b(x)
v(a(a(x))) -> u(v(x))
v(a(c(x))) -> u(b(d(x)))
v(c(x)) -> b(x)
w(a(a(x))) -> u(w(x))
w(a(c(x))) -> u(b(d(x)))
w(c(x)) -> b(x)





Using the Dependency Graph resulted in no new DP problems.

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