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

Innermost 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
         ↳Rewriting Transformation
       →DP Problem 2
         ↳Remaining


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


Strategy:

innermost




On this DP problem, a Rewriting 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
         ↳Rw
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



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




The following remains to be proven:


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




The following remains to be proven:

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