Term Rewriting System R:
[x, y]
a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) -> A(b(a(b(a(x)))))
A(a(f(x, y))) -> A(b(a(x)))
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> A(b(a(b(a(y)))))
A(a(f(x, y))) -> A(b(a(y)))
A(a(f(x, y))) -> A(y)
F(a(x), a(y)) -> A(f(x, y))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
F(a(x), a(y)) -> A(f(x, y))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost




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

F(a(x), a(y)) -> A(f(x, y))
two new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(b(x'')), a(b(y''))) -> A(b(f(x'', y'')))

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

A(a(f(x, y))) -> A(y)
A(a(f(x, y))) -> A(x)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(x), a(y)) -> F(x, y)
F(b(x), b(y)) -> F(x, y)


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 3
Forward Instantiation Transformation


Dependency Pairs:

A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) -> A(y)


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost




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

A(a(f(x, y))) -> A(y)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
F(a(x), a(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost




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

F(a(x), a(y)) -> F(x, y)
three new Dependency Pairs are created:

F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))

The transformation is resulting in one new DP problem:



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


Dependency Pairs:

F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(b(x), b(y)) -> F(x, y)
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost




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

F(b(x), b(y)) -> F(x, y)
four new Dependency Pairs are created:

F(b(b(x'')), b(b(y''))) -> F(b(x''), b(y''))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
FwdInst
             ...
               →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

A(a(f(x, a(f(a(f(x'''', y'''')), y''))))) -> A(a(f(a(f(x'''', y'''')), y'')))
A(a(f(x, a(f(x'', y''))))) -> A(a(f(x'', y'')))
A(a(f(a(f(x'', y'')), y))) -> A(a(f(x'', y'')))
F(b(a(a(a(x'''''')))), b(a(a(a(y''''''))))) -> F(a(a(a(x''''''))), a(a(a(y''''''))))
F(b(a(b(x''''))), b(a(b(y'''')))) -> F(a(b(x'''')), a(b(y'''')))
F(b(a(a(x''''))), b(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))
F(b(b(x'')), b(b(y''))) -> F(b(x''), b(y''))
F(a(b(x'')), a(b(y''))) -> F(b(x''), b(y''))
F(a(a(x'')), a(a(y''))) -> F(a(x''), a(y''))
A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
F(a(a(x'')), a(a(y''))) -> A(a(f(x'', y'')))
F(a(a(a(x''''))), a(a(a(y'''')))) -> F(a(a(x'''')), a(a(y'''')))


Rules:


a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) -> a(f(x, y))
f(b(x), b(y)) -> b(f(x, y))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes