Term Rewriting System R:
[x, y, z]
f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

F(x, g(y)) -> F(h(x), i(x, y))
F(x, g(y)) -> I(x, y)
I(x, j(y, z)) -> J(g(y), i(x, z))
I(x, j(y, z)) -> I(x, z)
I(h(x), j(j(y, z), 0)) -> J(i(h(x), j(y, z)), i(x, j(y, z)))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
J(g(x), g(y)) -> J(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
Nar


Dependency Pair:

J(g(x), g(y)) -> J(x, y)


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

J(g(x), g(y)) -> J(x, y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1))=  1 + x1  
  POL(J(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
J(x1, x2) -> J(x1, x2)
g(x1) -> g(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 4
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
Nar


Dependency Pair:


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))


The following usable rule for innermost can be oriented:

j(g(x), g(y)) -> g(j(x, y))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1, x2))=  1 + x1 + x2  
  POL(0)=  1  
  POL(g(x1))=  x1  
  POL(h(x1))=  1 + x1  
  POL(j(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
I(x1, x2) -> I(x1, x2)
h(x1) -> h(x1)
j(x1, x2) -> j(x1, x2)
g(x1) -> g(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pair:

I(x, j(y, z)) -> I(x, z)


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

I(x, j(y, z)) -> I(x, z)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(I(x1, x2))=  x1 + x2  
  POL(j(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
I(x1, x2) -> I(x1, x2)
j(x1, x2) -> j(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 5
AFS
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Narrowing Transformation


Dependency Pair:

F(x, g(y)) -> F(h(x), i(x, y))


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




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

F(x, g(y)) -> F(h(x), i(x, y))
three new Dependency Pairs are created:

F(x'', g(j(0, 0))) -> F(h(x''), g(0))
F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))
F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 7
Instantiation Transformation


Dependency Pairs:

F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))
F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




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

F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))
two new Dependency Pairs are created:

F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 7
Inst
             ...
               →DP Problem 8
Instantiation Transformation


Dependency Pairs:

F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))
F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost




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

F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))
three new Dependency Pairs are created:

F(h(h(x'''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''))), j(i(h(h(x'''')), j(y''', z'')), i(h(x''''), j(y''', z''))))
F(h(h(x'''''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''''))), j(i(h(h(x'''''')), j(y''', z'')), i(h(x''''''), j(y''', z''))))
F(h(h(h(x''''''))), g(j(j(y''', z''), 0))) -> F(h(h(h(h(x'''''')))), j(i(h(h(h(x''''''))), j(y''', z'')), i(h(h(x'''''')), j(y''', z''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 7
Inst
             ...
               →DP Problem 9
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

F(h(h(h(x''''''))), g(j(j(y''', z''), 0))) -> F(h(h(h(h(x'''''')))), j(i(h(h(h(x''''''))), j(y''', z'')), i(h(h(x'''''')), j(y''', z''))))
F(h(h(x'''''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''''))), j(i(h(h(x'''''')), j(y''', z'')), i(h(x''''''), j(y''', z''))))
F(h(h(x'''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''))), j(i(h(h(x'''')), j(y''', z'')), i(h(x''''), j(y''', z''))))
F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))


Rules:


f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:09 minutes