Term Rewriting System R:
[x]
a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

A(f, 0) -> A(s, 0)
A(d, a(s, x)) -> A(s, a(s, a(d, a(p, a(s, x)))))
A(d, a(s, x)) -> A(s, a(d, a(p, a(s, x))))
A(d, a(s, x)) -> A(d, a(p, a(s, x)))
A(d, a(s, x)) -> A(p, a(s, x))
A(f, a(s, x)) -> A(d, a(f, a(p, a(s, x))))
A(f, a(s, x)) -> A(f, a(p, a(s, x)))
A(f, a(s, x)) -> A(p, a(s, x))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

A(d, a(s, x)) -> A(d, a(p, a(s, x)))


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





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

A(d, a(s, x)) -> A(d, a(p, a(s, x)))
one new Dependency Pair is created:

A(d, a(s, x'')) -> A(d, x'')

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

A(d, a(s, x'')) -> A(d, x'')


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





The following dependency pair can be strictly oriented:

A(d, a(s, x'')) -> A(d, 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(d)=  0  
  POL(s)=  0  
  POL(a(x1, x2))=  1 + x2  
  POL(A(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

A(f, a(s, x)) -> A(f, a(p, a(s, x)))


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





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

A(f, a(s, x)) -> A(f, a(p, a(s, x)))
one new Dependency Pair is created:

A(f, a(s, x'')) -> A(f, x'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Nar
           →DP Problem 5
             ↳Polynomial Ordering


Dependency Pair:

A(f, a(s, x'')) -> A(f, x'')


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





The following dependency pair can be strictly oriented:

A(f, a(s, x'')) -> A(f, 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(s)=  0  
  POL(a(x1, x2))=  1 + x2  
  POL(A(x1, x2))=  x2  
  POL(f)=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Nar
           →DP Problem 5
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x





Using the Dependency Graph resulted in no new DP problems.

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