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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

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

A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(u(x''))) -> B(a(x''))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(x)) -> A(a(x))
C(a(x)) -> A(c(c(x)))
B(c(x)) -> C(b(b(x)))
B(c(x)) -> B(x)


Rules:


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


Strategy:

innermost




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

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

A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(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
                 ↳Narrowing Transformation


Dependency Pairs:

A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
C(a(x)) -> C(c(x))
B(c(x)) -> C(b(b(x)))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(x)) -> A(c(c(x)))
C(a(x)) -> C(x)


Rules:


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


Strategy:

innermost




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

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

B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(v(x''))) -> C(b(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Narrowing Transformation


Dependency Pairs:

A(b(b(x''))) -> A(b(a(a(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(x)) -> B(b(x))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')


Rules:


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


Strategy:

innermost




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

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

B(c(c(x''))) -> B(c(b(b(x''))))
B(c(v(x''))) -> B(x'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Narrowing Transformation


Dependency Pairs:

A(b(u(x''))) -> A(x'')
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(x)) -> A(c(c(x)))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(b(x''))) -> A(b(a(a(x''))))


Rules:


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


Strategy:

innermost




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

C(a(x)) -> A(c(c(x)))
two new Dependency Pairs are created:

C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(w(x''))) -> A(c(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 6
                 ↳Narrowing Transformation


Dependency Pairs:

B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(b(x''))) -> A(b(a(a(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
C(a(x)) -> C(c(x))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
A(b(u(x''))) -> A(x'')


Rules:


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


Strategy:

innermost




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

C(a(x)) -> C(c(x))
two new Dependency Pairs are created:

C(a(a(x''))) -> C(a(c(c(x''))))
C(a(w(x''))) -> C(x'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 7
                 ↳Forward Instantiation Transformation


Dependency Pairs:

C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(x)) -> A(x)
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
B(c(v(x''))) -> B(x'')


Rules:


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


Strategy:

innermost




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

A(b(x)) -> A(x)
three new Dependency Pairs are created:

A(b(b(x''))) -> A(b(x''))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 8
                 ↳Forward Instantiation Transformation


Dependency Pairs:

A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(a(x''))) -> C(a(c(c(x''))))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(x)) -> B(x)
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
C(a(w(x''))) -> C(x'')


Rules:


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


Strategy:

innermost




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

B(c(x)) -> B(x)
three new Dependency Pairs are created:

B(c(c(x''))) -> B(c(x''))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(v(x'''')))) -> B(c(v(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 9
                 ↳Forward Instantiation Transformation


Dependency Pairs:

B(c(c(v(x'''')))) -> B(c(v(x'''')))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(x)) -> C(x)
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(u(x''))) -> B(a(x''))
A(b(b(u(x'''')))) -> A(b(u(x'''')))


Rules:


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


Strategy:

innermost




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

C(a(x)) -> C(x)
three new Dependency Pairs are created:

C(a(a(x''))) -> C(a(x''))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(w(x'''')))) -> C(a(w(x'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 10
                 ↳Forward Instantiation Transformation


Dependency Pairs:

C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(u(x''))) -> A(x'')
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
B(c(v(x''))) -> C(b(x''))
B(c(c(v(x'''')))) -> B(c(v(x'''')))


Rules:


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


Strategy:

innermost




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

A(b(u(x''))) -> A(x'')
four new Dependency Pairs are created:

A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 11
                 ↳Forward Instantiation Transformation


Dependency Pairs:

A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(v(x'''')))) -> B(c(v(x'''')))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(v(x''))) -> B(x'')
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(w(x''))) -> A(c(x''))
C(a(a(w(x'''')))) -> C(a(w(x'''')))


Rules:


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


Strategy:

innermost




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

B(c(v(x''))) -> B(x'')
four new Dependency Pairs are created:

B(c(v(c(c(x''''))))) -> B(c(c(x'''')))
B(c(v(c(v(x''''))))) -> B(c(v(x'''')))
B(c(v(c(c(c(x'''''')))))) -> B(c(c(c(x''''''))))
B(c(v(c(c(v(x'''''')))))) -> B(c(c(v(x''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 12
                 ↳Forward Instantiation Transformation


Dependency Pairs:

B(c(v(c(c(v(x'''''')))))) -> B(c(c(v(x''''''))))
B(c(v(c(c(c(x'''''')))))) -> B(c(c(c(x''''''))))
B(c(v(c(v(x''''))))) -> B(c(v(x'''')))
B(c(v(c(c(x''''))))) -> B(c(c(x'''')))
B(c(c(v(x'''')))) -> B(c(v(x'''')))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(w(x''))) -> C(x'')
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(u(x''))) -> B(a(x''))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))


Rules:


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


Strategy:

innermost




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

C(a(w(x''))) -> C(x'')
four new Dependency Pairs are created:

C(a(w(a(a(x''''))))) -> C(a(a(x'''')))
C(a(w(a(w(x''''))))) -> C(a(w(x'''')))
C(a(w(a(a(a(x'''''')))))) -> C(a(a(a(x''''''))))
C(a(w(a(a(w(x'''''')))))) -> C(a(a(w(x''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 13
                 ↳Polynomial Ordering


Dependency Pairs:

C(a(w(a(a(w(x'''''')))))) -> C(a(a(w(x''''''))))
C(a(w(a(a(a(x'''''')))))) -> C(a(a(a(x''''''))))
C(a(w(a(w(x''''))))) -> C(a(w(x'''')))
C(a(w(a(a(x''''))))) -> C(a(a(x'''')))
C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(v(c(c(c(x'''''')))))) -> B(c(c(c(x''''''))))
B(c(v(c(v(x''''))))) -> B(c(v(x'''')))
B(c(v(c(c(x''''))))) -> B(c(c(x'''')))
B(c(c(v(x'''')))) -> B(c(v(x'''')))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(v(x''))) -> C(b(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
B(c(v(c(c(v(x'''''')))))) -> B(c(c(v(x''''''))))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

B(c(v(c(c(c(x'''''')))))) -> B(c(c(c(x''''''))))
B(c(v(c(v(x''''))))) -> B(c(v(x'''')))
B(c(v(c(c(x''''))))) -> B(c(c(x'''')))
B(c(v(x''))) -> C(b(x''))
B(c(v(c(c(v(x'''''')))))) -> B(c(c(v(x''''''))))


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(C(x1))=  x1  
  POL(v(x1))=  1 + x1  
  POL(B(x1))=  x1  
  POL(b(x1))=  x1  
  POL(a(x1))=  x1  
  POL(w(x1))=  x1  
  POL(u(x1))=  x1  
  POL(A(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 14
                 ↳Dependency Graph


Dependency Pairs:

C(a(w(a(a(w(x'''''')))))) -> C(a(a(w(x''''''))))
C(a(w(a(a(a(x'''''')))))) -> C(a(a(a(x''''''))))
C(a(w(a(w(x''''))))) -> C(a(w(x'''')))
C(a(w(a(a(x''''))))) -> C(a(a(x'''')))
C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(v(x'''')))) -> B(c(v(x'''')))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 15
                 ↳Polynomial Ordering


Dependency Pairs:

C(a(w(a(a(a(x'''''')))))) -> C(a(a(a(x''''''))))
C(a(w(a(w(x''''))))) -> C(a(w(x'''')))
C(a(w(a(a(x''''))))) -> C(a(a(x'''')))
C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
C(a(w(x''))) -> A(c(x''))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(w(a(a(w(x'''''')))))) -> C(a(a(w(x''''''))))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

C(a(w(a(a(a(x'''''')))))) -> C(a(a(a(x''''''))))
C(a(w(a(w(x''''))))) -> C(a(w(x'''')))
C(a(w(a(a(x''''))))) -> C(a(a(x'''')))
C(a(w(x''))) -> A(c(x''))
C(a(w(a(a(w(x'''''')))))) -> C(a(a(w(x''''''))))


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(C(x1))=  x1  
  POL(v(x1))=  1 + x1  
  POL(B(x1))=  x1  
  POL(b(x1))=  x1  
  POL(a(x1))=  x1  
  POL(w(x1))=  1 + x1  
  POL(u(x1))=  x1  
  POL(A(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 16
                 ↳Dependency Graph


Dependency Pairs:

C(a(a(w(x'''')))) -> C(a(w(x'''')))
C(a(a(a(x'''')))) -> C(a(a(x'''')))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 17
                 ↳Polynomial Ordering


Dependency Pairs:

A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
A(b(u(x''))) -> B(a(x''))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(a(a(x'''')))) -> C(a(a(x'''')))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

A(b(u(b(b(u(x'''''')))))) -> A(b(b(u(x''''''))))
A(b(u(b(b(b(x'''''')))))) -> A(b(b(b(x''''''))))
A(b(u(b(u(x''''))))) -> A(b(u(x'''')))
A(b(u(b(b(x''''))))) -> A(b(b(x'''')))
A(b(u(x''))) -> B(a(x''))


Additionally, the following usable rules for innermost can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(C(x1))=  x1  
  POL(v(x1))=  1 + x1  
  POL(B(x1))=  x1  
  POL(b(x1))=  x1  
  POL(a(x1))=  x1  
  POL(w(x1))=  1 + x1  
  POL(u(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 18
                 ↳Dependency Graph


Dependency Pairs:

A(b(b(u(x'''')))) -> A(b(u(x'''')))
A(b(b(b(x'''')))) -> A(b(b(x'''')))
A(b(b(x''))) -> A(b(x''))
A(b(b(x''))) -> A(b(a(a(x''))))
B(c(c(c(x'''')))) -> B(c(c(x'''')))
B(c(c(x''))) -> B(c(x''))
B(c(c(x''))) -> B(c(b(b(x''))))
C(a(a(x''))) -> C(a(x''))
C(a(a(x''))) -> C(a(c(c(x''))))
B(c(c(x''))) -> C(b(c(b(b(x'')))))
A(b(b(x''))) -> B(a(b(a(a(x'')))))
C(a(a(x''))) -> A(c(a(c(c(x'')))))
C(a(a(a(x'''')))) -> C(a(a(x'''')))


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 19
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:08 minutes