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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

A(d(x)) -> B(a(x))
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
B(c(x)) -> B(x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

B(c(x)) -> B(x)
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
A(d(x)) -> B(a(x))


Rules:


a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

A(d(x)) -> B(a(x))
two new Dependency Pairs are created:

A(d(d(x''))) -> B(d(c(b(a(x'')))))
A(d(c(x''))) -> B(x'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

A(d(c(x''))) -> B(x'')
A(d(x)) -> A(x)
B(c(x)) -> A(b(x))
B(c(x)) -> B(x)


Rules:


a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

B(c(x)) -> A(b(x))
two new Dependency Pairs are created:

B(c(c(x''))) -> A(c(d(a(b(x'')))))
B(c(d(x''))) -> A(x'')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

A(d(x)) -> A(x)
B(c(d(x''))) -> A(x'')
B(c(x)) -> B(x)
A(d(c(x''))) -> B(x'')


Rules:


a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x





The following dependency pairs can be strictly oriented:

A(d(x)) -> A(x)
B(c(d(x''))) -> A(x'')
A(d(c(x''))) -> B(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(c(x1))=  x1  
  POL(B(x1))=  x1  
  POL(d(x1))=  1 + x1  
  POL(A(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

B(c(x)) -> B(x)


Rules:


a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x





The following dependency pair can be strictly oriented:

B(c(x)) -> B(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(c(x1))=  1 + x1  
  POL(B(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


a(d(x)) -> d(c(b(a(x))))
a(c(x)) -> x
b(c(x)) -> c(d(a(b(x))))
b(d(x)) -> x





Using the Dependency Graph resulted in no new DP problems.

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