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

Innermost 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


Strategy:

innermost




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


Strategy:

innermost




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
Forward Instantiation Transformation


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 4
Forward Instantiation Transformation


Dependency Pairs:

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


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)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 5
Forward Instantiation Transformation


Dependency Pairs:

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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 6
Forward Instantiation Transformation


Dependency Pairs:

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


Strategy:

innermost




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

B(c(d(x''))) -> A(x'')
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 7
Polynomial Ordering


Dependency Pairs:

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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost 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  
  POL(d(x1))=  x1  
  POL(A(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 8
Dependency Graph


Dependency Pairs:

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


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 9
Polynomial Ordering


Dependency Pair:

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(d(x1))=  1 + x1  
  POL(A(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 10
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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes