Term Rewriting System R:
[x, y, z]
f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(0), y, z) -> F(0, s(y), s(z))
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(0, s(s(y)), s(0)) -> F(0, y, s(0))
F(0, s(0), s(s(z))) -> F(0, s(0), z)
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y)), s(0)) -> F(0, y, s(0))
one new Dependency Pair is created:

F(0, s(s(s(s(y'')))), s(0)) -> F(0, s(s(y'')), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(s(s(s(y'')))), s(0)) -> F(0, s(s(y'')), s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(s(s(y'')))), s(0)) -> F(0, s(s(y'')), s(0))
one new Dependency Pair is created:

F(0, s(s(s(s(s(s(y'''')))))), s(0)) -> F(0, s(s(s(s(y'''')))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 6
Polynomial Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(s(s(s(s(s(y'''')))))), s(0)) -> F(0, s(s(s(s(y'''')))), s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(0, s(s(s(s(s(s(y'''')))))), s(0)) -> F(0, s(s(s(s(y'''')))), s(0))


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(0)=  1  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 5
FwdInst
             ...
               →DP Problem 7
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Forward Instantiation Transformation
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(0), s(s(z))) -> F(0, s(0), z)


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(0), s(s(z))) -> F(0, s(0), z)
one new Dependency Pair is created:

F(0, s(0), s(s(s(s(z''))))) -> F(0, s(0), s(s(z'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(0), s(s(s(s(z''))))) -> F(0, s(0), s(s(z'')))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(0), s(s(s(s(z''))))) -> F(0, s(0), s(s(z'')))
one new Dependency Pair is created:

F(0, s(0), s(s(s(s(s(s(z''''))))))) -> F(0, s(0), s(s(s(s(z'''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 9
Polynomial Ordering
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:

F(0, s(0), s(s(s(s(s(s(z''''))))))) -> F(0, s(0), s(s(s(s(z'''')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

F(0, s(0), s(s(s(s(s(s(z''''))))))) -> F(0, s(0), s(s(s(s(z'''')))))


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(0)=  1  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 8
FwdInst
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
Nar
       →DP Problem 4
Nar


Dependency Pair:


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Narrowing Transformation
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))
two new Dependency Pairs are created:

F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Narrowing Transformation
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))
F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
three new Dependency Pairs are created:

F(0, s(s(0)), s(s(0))) -> F(0, 0, s(s(0)))
F(0, s(s(s(0))), s(s(0))) -> F(0, s(0), s(s(0)))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 12
Narrowing Transformation
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))
three new Dependency Pairs are created:

F(0, s(s(0)), s(s(s(z''')))) -> F(0, 0, s(f(0, s(s(0)), s(z'''))))
F(0, s(s(y''')), s(s(s(0)))) -> F(0, y''', f(0, y''', f(0, y''', s(0))))
F(0, s(s(y''')), s(s(s(s(z'))))) -> F(0, y''', f(0, y''', f(0, y''', f(0, s(s(y''')), s(z')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 13
Forward Instantiation Transformation
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y''')), s(s(s(s(z'))))) -> F(0, y''', f(0, y''', f(0, y''', f(0, s(s(y''')), s(z')))))
F(0, s(s(y''')), s(s(s(0)))) -> F(0, y''', f(0, y''', f(0, y''', s(0))))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
four new Dependency Pairs are created:

F(0, s(s(y'')), s(s(s(z'')))) -> F(0, s(s(y'')), s(s(z'')))
F(0, s(s(s(s(y''')))), s(s(s(0)))) -> F(0, s(s(s(s(y''')))), s(s(0)))
F(0, s(s(y')), s(s(s(s(0))))) -> F(0, s(s(y')), s(s(s(0))))
F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 14
Forward Instantiation Transformation
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))
F(0, s(s(y')), s(s(s(s(0))))) -> F(0, s(s(y')), s(s(s(0))))
F(0, s(s(s(s(y''')))), s(s(s(0)))) -> F(0, s(s(s(s(y''')))), s(s(0)))
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, s(s(y'')), s(s(z'')))
F(0, s(s(y''')), s(s(s(0)))) -> F(0, y''', f(0, y''', f(0, y''', s(0))))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))
F(0, s(s(y''')), s(s(s(s(z'))))) -> F(0, y''', f(0, y''', f(0, y''', f(0, s(s(y''')), s(z')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(0, s(s(y'')), s(s(s(z'')))) -> F(0, s(s(y'')), s(s(z'')))
seven new Dependency Pairs are created:

F(0, s(s(s(s(y'''')))), s(s(s(0)))) -> F(0, s(s(s(s(y'''')))), s(s(0)))
F(0, s(s(y''')), s(s(s(s(0))))) -> F(0, s(s(y''')), s(s(s(0))))
F(0, s(s(y''')), s(s(s(s(s(z'''')))))) -> F(0, s(s(y''')), s(s(s(s(z'''')))))
F(0, s(s(y'''')), s(s(s(s(z''''))))) -> F(0, s(s(y'''')), s(s(s(z''''))))
F(0, s(s(s(s(y''''')))), s(s(s(s(0))))) -> F(0, s(s(s(s(y''''')))), s(s(s(0))))
F(0, s(s(y'''')), s(s(s(s(s(0)))))) -> F(0, s(s(y'''')), s(s(s(s(0)))))
F(0, s(s(y'''')), s(s(s(s(s(s(z'''''))))))) -> F(0, s(s(y'''')), s(s(s(s(s(z'''''))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 15
Polynomial Ordering
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y'''')), s(s(s(s(s(s(z'''''))))))) -> F(0, s(s(y'''')), s(s(s(s(s(z'''''))))))
F(0, s(s(y'''')), s(s(s(s(s(0)))))) -> F(0, s(s(y'''')), s(s(s(s(0)))))
F(0, s(s(s(s(y''''')))), s(s(s(s(0))))) -> F(0, s(s(s(s(y''''')))), s(s(s(0))))
F(0, s(s(y'''')), s(s(s(s(z''''))))) -> F(0, s(s(y'''')), s(s(s(z''''))))
F(0, s(s(y''')), s(s(s(s(s(z'''')))))) -> F(0, s(s(y''')), s(s(s(s(z'''')))))
F(0, s(s(y''')), s(s(s(s(0))))) -> F(0, s(s(y''')), s(s(s(0))))
F(0, s(s(s(s(y'''')))), s(s(s(0)))) -> F(0, s(s(s(s(y'''')))), s(s(0)))
F(0, s(s(y')), s(s(s(s(0))))) -> F(0, s(s(y')), s(s(s(0))))
F(0, s(s(s(s(y''')))), s(s(s(0)))) -> F(0, s(s(s(s(y''')))), s(s(0)))
F(0, s(s(y''')), s(s(s(0)))) -> F(0, y''', f(0, y''', f(0, y''', s(0))))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))
F(0, s(s(y''')), s(s(s(s(z'))))) -> F(0, y''', f(0, y''', f(0, y''', f(0, s(s(y''')), s(z')))))
F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(0, s(s(y''')), s(s(s(0)))) -> F(0, y''', f(0, y''', f(0, y''', s(0))))
F(0, s(s(s(s(y')))), s(s(0))) -> F(0, s(s(y')), f(0, y', s(0)))
F(0, s(s(y''')), s(s(s(s(z'))))) -> F(0, y''', f(0, y''', f(0, y''', f(0, s(s(y''')), s(z')))))


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(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(f(x1, x2, x3))=  0  
  POL(F(x1, x2, x3))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 16
Dependency Graph
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y'''')), s(s(s(s(s(s(z'''''))))))) -> F(0, s(s(y'''')), s(s(s(s(s(z'''''))))))
F(0, s(s(y'''')), s(s(s(s(s(0)))))) -> F(0, s(s(y'''')), s(s(s(s(0)))))
F(0, s(s(s(s(y''''')))), s(s(s(s(0))))) -> F(0, s(s(s(s(y''''')))), s(s(s(0))))
F(0, s(s(y'''')), s(s(s(s(z''''))))) -> F(0, s(s(y'''')), s(s(s(z''''))))
F(0, s(s(y''')), s(s(s(s(s(z'''')))))) -> F(0, s(s(y''')), s(s(s(s(z'''')))))
F(0, s(s(y''')), s(s(s(s(0))))) -> F(0, s(s(y''')), s(s(s(0))))
F(0, s(s(s(s(y'''')))), s(s(s(0)))) -> F(0, s(s(s(s(y'''')))), s(s(0)))
F(0, s(s(y')), s(s(s(s(0))))) -> F(0, s(s(y')), s(s(s(0))))
F(0, s(s(s(s(y''')))), s(s(s(0)))) -> F(0, s(s(s(s(y''')))), s(s(0)))
F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 17
Polynomial Ordering
       →DP Problem 4
Nar


Dependency Pairs:

F(0, s(s(y'''')), s(s(s(s(s(0)))))) -> F(0, s(s(y'''')), s(s(s(s(0)))))
F(0, s(s(y'''')), s(s(s(s(z''''))))) -> F(0, s(s(y'''')), s(s(s(z''''))))
F(0, s(s(y''')), s(s(s(s(s(z'''')))))) -> F(0, s(s(y''')), s(s(s(s(z'''')))))
F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))
F(0, s(s(y'''')), s(s(s(s(s(s(z'''''))))))) -> F(0, s(s(y'''')), s(s(s(s(s(z'''''))))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(0, s(s(y'''')), s(s(s(s(s(0)))))) -> F(0, s(s(y'''')), s(s(s(s(0)))))
F(0, s(s(y'''')), s(s(s(s(z''''))))) -> F(0, s(s(y'''')), s(s(s(z''''))))
F(0, s(s(y''')), s(s(s(s(s(z'''')))))) -> F(0, s(s(y''')), s(s(s(s(z'''')))))
F(0, s(s(y')), s(s(s(s(s(z''')))))) -> F(0, s(s(y')), s(s(s(s(z''')))))
F(0, s(s(y'''')), s(s(s(s(s(s(z'''''))))))) -> F(0, s(s(y'''')), s(s(s(s(s(z'''''))))))


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(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 18
Dependency Graph
       →DP Problem 4
Nar


Dependency Pair:


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Narrowing Transformation


Dependency Pairs:

F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
three new Dependency Pairs are created:

F(s(0), s(y''), s(z'')) -> F(0, y'', f(0, s(s(y'')), s(z'')))
F(s(x''), s(y''), s(0)) -> F(x'', y'', f(x'', y'', s(0)))
F(s(x''), s(y''), s(s(z''))) -> F(x'', y'', f(x'', y'', f(s(x''), s(y''), z'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Narrowing Transformation


Dependency Pairs:

F(s(x''), s(y''), s(s(z''))) -> F(x'', y'', f(x'', y'', f(s(x''), s(y''), z'')))
F(s(x''), s(y''), s(0)) -> F(x'', y'', f(x'', y'', s(0)))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x''), s(y''), s(0)) -> F(x'', y'', f(x'', y'', s(0)))
six new Dependency Pairs are created:

F(s(0), s(0), s(0)) -> F(0, 0, s(s(0)))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(0), s(s(0)), s(0)) -> F(0, s(0), s(s(0)))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(0), s(s(s(y'))), s(0)) -> F(0, s(s(y')), f(0, y', s(0)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 20
Narrowing Transformation


Dependency Pairs:

F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(x''), s(y''), s(s(z''))) -> F(x'', y'', f(x'', y'', f(s(x''), s(y''), z'')))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x''), s(y''), s(s(z''))) -> F(x'', y'', f(x'', y'', f(s(x''), s(y''), z'')))
five new Dependency Pairs are created:

F(s(0), s(0), s(s(z'''))) -> F(0, 0, s(f(s(0), s(0), z''')))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(0), s(y'''), s(s(z'''))) -> F(0, y''', f(0, y''', f(0, s(s(y''')), s(z'''))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 21
Forward Instantiation Transformation


Dependency Pairs:

F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x), s(y), 0) -> F(x, y, s(0))
five new Dependency Pairs are created:

F(s(s(x'')), s(0), 0) -> F(s(x''), 0, s(0))
F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 22
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
F(s(s(x'')), s(0), 0) -> F(s(x''), 0, s(0))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x), 0, s(z)) -> F(x, s(0), z)
seven new Dependency Pairs are created:

F(s(s(x'')), 0, s(s(z''))) -> F(s(x''), s(0), s(z''))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(s(x''''))), 0, s(0)) -> F(s(s(x'''')), s(0), 0)

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 23
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(x''''))), 0, s(0)) -> F(s(s(x'''')), s(0), 0)
F(s(s(x'')), s(0), 0) -> F(s(x''), 0, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(x'')), s(0), 0) -> F(s(x''), 0, s(0))
one new Dependency Pair is created:

F(s(s(s(s(x'''''')))), s(0), 0) -> F(s(s(s(x''''''))), 0, s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 25
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(s(s(x'''''')))), s(0), 0) -> F(s(s(s(x''''''))), 0, s(0))
F(s(s(s(x''''))), 0, s(0)) -> F(s(s(x'''')), s(0), 0)


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(s(x''''))), 0, s(0)) -> F(s(s(x'''')), s(0), 0)
one new Dependency Pair is created:

F(s(s(s(s(s(x''''''''))))), 0, s(0)) -> F(s(s(s(s(x'''''''')))), s(0), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 27
Polynomial Ordering


Dependency Pairs:

F(s(s(s(s(s(x''''''''))))), 0, s(0)) -> F(s(s(s(s(x'''''''')))), s(0), 0)
F(s(s(s(s(x'''''')))), s(0), 0) -> F(s(s(s(x''''''))), 0, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(s(s(s(s(s(x''''''''))))), 0, s(0)) -> F(s(s(s(s(x'''''''')))), s(0), 0)
F(s(s(s(s(x'''''')))), s(0), 0) -> F(s(s(s(x''''''))), 0, s(0))


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(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 32
Dependency Graph


Dependency Pair:


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 24
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x'')), 0, s(s(z''))) -> F(s(x''), s(0), s(z''))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
11 new Dependency Pairs are created:

F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 26
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
F(s(s(x'')), 0, s(s(z''))) -> F(s(x''), s(0), s(z''))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(x'')), s(s(y'')), 0) -> F(s(x''), s(y''), s(0))
seven new Dependency Pairs are created:

F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 28
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
F(s(s(x'')), 0, s(s(z''))) -> F(s(x''), s(0), s(z''))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(x'')), 0, s(s(z''))) -> F(s(x''), s(0), s(z''))
11 new Dependency Pairs are created:

F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(s(x''''))), 0, s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(x''')), 0, s(s(s(0)))) -> F(s(x'''), s(0), s(s(0)))
F(s(s(x''')), 0, s(s(s(s(z''''))))) -> F(s(x'''), s(0), s(s(s(z''''))))
F(s(s(x'''')), 0, s(s(s(z'''')))) -> F(s(x''''), s(0), s(s(z'''')))
F(s(s(s(0))), 0, s(s(s(0)))) -> F(s(s(0)), s(0), s(s(0)))
F(s(s(s(x'''''))), 0, s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(s(z'''''''))))) -> F(s(s(0)), s(0), s(s(s(z'''''''))))
F(s(s(x'''')), 0, s(s(s(s(0))))) -> F(s(x''''), s(0), s(s(s(0))))
F(s(s(x'''')), 0, s(s(s(s(s(z''''')))))) -> F(s(x''''), s(0), s(s(s(s(z''''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 29
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(x'''')), 0, s(s(s(s(s(z''''')))))) -> F(s(x''''), s(0), s(s(s(s(z''''')))))
F(s(s(x'''')), 0, s(s(s(s(0))))) -> F(s(x''''), s(0), s(s(s(0))))
F(s(s(s(0))), 0, s(s(s(s(z'''''''))))) -> F(s(s(0)), s(0), s(s(s(z'''''''))))
F(s(s(s(x'''''))), 0, s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(0)))) -> F(s(s(0)), s(0), s(s(0)))
F(s(s(x'''')), 0, s(s(s(z'''')))) -> F(s(x''''), s(0), s(s(z'''')))
F(s(s(x''')), 0, s(s(s(s(z''''))))) -> F(s(x'''), s(0), s(s(s(z''''))))
F(s(s(x''')), 0, s(s(s(0)))) -> F(s(x'''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x''''))), 0, s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(x''), s(y''), s(s(z''))) -> F(s(x''), s(y''), s(z''))
17 new Dependency Pairs are created:

F(s(s(0)), s(y'''), s(s(0))) -> F(s(s(0)), s(y'''), s(0))
F(s(s(x'''')), s(0), s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(x'''')), s(s(y'''')), s(s(0))) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'''), s(y'''), s(s(s(0)))) -> F(s(x'''), s(y'''), s(s(0)))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(s(0)), s(y''''), s(s(s(0)))) -> F(s(s(0)), s(y''''), s(s(0)))
F(s(s(x''''')), s(0), s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(x''''')), s(s(y''''')), s(s(s(0)))) -> F(s(s(x''''')), s(s(y''''')), s(s(0)))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(s(x'''''')), s(s(y'''''')), s(s(0))) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(0))), s(s(y''''''''')), s(s(0))) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x'''''''))), s(s(0)), s(s(0))) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(s(0))) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 30
Forward Instantiation Transformation


Dependency Pairs:

F(s(s(x'''')), 0, s(s(s(s(0))))) -> F(s(x''''), s(0), s(s(s(0))))
F(s(s(s(0))), 0, s(s(s(s(z'''''''))))) -> F(s(s(0)), s(0), s(s(s(z'''''''))))
F(s(s(s(x'''''))), 0, s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(0)))) -> F(s(s(0)), s(0), s(s(0)))
F(s(s(x'''')), 0, s(s(s(z'''')))) -> F(s(x''''), s(0), s(s(z'''')))
F(s(s(x''')), 0, s(s(s(s(z''''))))) -> F(s(x'''), s(0), s(s(s(z''''))))
F(s(s(x''')), 0, s(s(s(0)))) -> F(s(x'''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x''''))), 0, s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(s(x''''')), s(s(y''''')), s(s(s(0)))) -> F(s(s(x''''')), s(s(y''''')), s(s(0)))
F(s(s(x''''')), s(0), s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(0)), s(y''''), s(s(s(0)))) -> F(s(s(0)), s(y''''), s(s(0)))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x'''), s(y'''), s(s(s(0)))) -> F(s(x'''), s(y'''), s(s(0)))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(s(0))) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(x'''''''))), s(s(0)), s(s(0))) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(0))), s(s(y''''''''')), s(s(0))) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(x'''''')), s(s(y'''''')), s(s(0))) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(x'''')), s(s(y'''')), s(s(0))) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(x'''')), s(0), s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(0)), s(y'''), s(s(0))) -> F(s(s(0)), s(y'''), s(0))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x'''')), 0, s(s(s(s(s(z''''')))))) -> F(s(x''''), s(0), s(s(s(s(z''''')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(x'''')), s(s(y'''')), s(0)) -> F(s(s(x'''')), s(s(y'''')), 0)
eight new Dependency Pairs are created:

F(s(s(s(0))), s(s(y''''')), s(0)) -> F(s(s(s(0))), s(s(y''''')), 0)
F(s(s(s(x''''''))), s(s(0)), s(0)) -> F(s(s(s(x''''''))), s(s(0)), 0)
F(s(s(s(x''''''))), s(s(s(y''''''))), s(0)) -> F(s(s(s(x''''''))), s(s(s(y''''''))), 0)
F(s(s(s(0))), s(s(y'''''')), s(0)) -> F(s(s(s(0))), s(s(y'''''')), 0)
F(s(s(s(x''''''''))), s(s(s(y''''''''))), s(0)) -> F(s(s(s(x''''''''))), s(s(s(y''''''''))), 0)
F(s(s(s(s(0)))), s(s(s(y'''''''''''))), s(0)) -> F(s(s(s(s(0)))), s(s(s(y'''''''''''))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(0))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(0))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 31
Polynomial Ordering


Dependency Pairs:

F(s(s(x'''')), 0, s(s(s(s(s(z''''')))))) -> F(s(x''''), s(0), s(s(s(s(z''''')))))
F(s(s(s(0))), 0, s(s(s(s(z'''''''))))) -> F(s(s(0)), s(0), s(s(s(z'''''''))))
F(s(s(s(x'''''))), 0, s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(0)))) -> F(s(s(0)), s(0), s(s(0)))
F(s(s(x'''')), 0, s(s(s(z'''')))) -> F(s(x''''), s(0), s(s(z'''')))
F(s(s(x''')), 0, s(s(s(s(z''''))))) -> F(s(x'''), s(0), s(s(s(z''''))))
F(s(s(x''')), 0, s(s(s(0)))) -> F(s(x'''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x''''))), 0, s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(s(x''''')), s(s(y''''')), s(s(s(0)))) -> F(s(s(x''''')), s(s(y''''')), s(s(0)))
F(s(s(x''''')), s(0), s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(0)), s(y''''), s(s(s(0)))) -> F(s(s(0)), s(y''''), s(s(0)))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x'''), s(y'''), s(s(s(0)))) -> F(s(x'''), s(y'''), s(s(0)))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(s(0))) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(x'''''''))), s(s(0)), s(s(0))) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(0))), s(s(y''''''''')), s(s(0))) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(x'''''')), s(s(y'''''')), s(s(0))) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(x'''')), s(s(y'''')), s(s(0))) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(x'''')), s(0), s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(0)), s(y'''), s(s(0))) -> F(s(s(0)), s(y'''), s(0))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(0))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(0))), 0)
F(s(s(s(s(0)))), s(s(s(y'''''''''''))), s(0)) -> F(s(s(s(s(0)))), s(s(s(y'''''''''''))), 0)
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(x''''''''))), s(s(s(y''''''''))), s(0)) -> F(s(s(s(x''''''''))), s(s(s(y''''''''))), 0)
F(s(s(s(0))), s(s(y'''''')), s(0)) -> F(s(s(s(0))), s(s(y'''''')), 0)
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), s(0)) -> F(s(s(s(x''''''))), s(s(s(y''''''))), 0)
F(s(s(s(x''''''))), s(s(0)), s(0)) -> F(s(s(s(x''''''))), s(s(0)), 0)
F(s(s(s(0))), s(s(y''''')), s(0)) -> F(s(s(s(0))), s(s(y''''')), 0)
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x'''')), 0, s(s(s(s(0))))) -> F(s(x''''), s(0), s(s(s(0))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(s(s(x'''')), 0, s(s(s(s(s(z''''')))))) -> F(s(x''''), s(0), s(s(s(s(z''''')))))
F(s(s(s(0))), 0, s(s(s(s(z'''''''))))) -> F(s(s(0)), s(0), s(s(s(z'''''''))))
F(s(s(s(x'''''))), 0, s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(0)))) -> F(s(s(0)), s(0), s(s(0)))
F(s(s(x'''')), 0, s(s(s(z'''')))) -> F(s(x''''), s(0), s(s(z'''')))
F(s(s(x''')), 0, s(s(s(s(z''''))))) -> F(s(x'''), s(0), s(s(s(z''''))))
F(s(s(x''')), 0, s(s(s(0)))) -> F(s(x'''), s(0), s(s(0)))
F(s(s(s(0))), 0, s(s(s(z''''')))) -> F(s(s(0)), s(0), s(s(z''''')))
F(s(s(s(x''''))), 0, s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), 0, s(s(0))) -> F(s(s(0)), s(0), s(0))
F(s(s(x''''')), 0, s(s(s(s(z'''))))) -> F(s(x'''''), s(0), s(s(s(z'''))))
F(s(s(x''''')), 0, s(s(s(0)))) -> F(s(x'''''), s(0), s(s(0)))
F(s(s(s(s(x''''''')))), s(s(s(s(y''''''')))), 0) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(s(x''''''')))), s(s(s(0))), 0) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(s(0)))), s(s(s(y'''''''''))), 0) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(s(x''''''))), s(s(s(y''''''))), 0) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(s(x''''))), s(s(s(y''''))), 0) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(s(x''''))), s(s(0)), 0) -> F(s(s(x'''')), s(0), s(0))
F(s(s(s(0))), s(s(y''')), 0) -> F(s(s(0)), s(y'''), s(0))
F(s(s(s(x'''))), s(s(s(y'''))), 0) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(s(x'''))), s(s(0)), 0) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(0))), s(s(y''''')), 0) -> F(s(s(0)), s(y'''''), s(0))
F(s(s(x')), s(s(y')), s(0)) -> F(s(x'), s(y'), f(x', y', f(s(x'), s(y'), 0)))
F(s(s(s(x'''))), 0, s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(x')), s(0), s(0)) -> F(s(x'), 0, f(x', s(0), 0))
F(s(x'''), s(y'''), s(s(s(z')))) -> F(x''', y''', f(x''', y''', f(x''', y''', f(s(x'''), s(y'''), z'))))
F(s(s(0)), s(y'''), s(0)) -> F(s(0), y''', f(0, s(y'''), s(s(0))))
F(s(x'''), s(y'''), s(s(0))) -> F(x''', y''', f(x''', y''', f(x''', y''', s(0))))
F(s(s(0)), s(y'''), s(s(z'''))) -> F(s(0), y''', f(0, s(y'''), s(f(s(s(0)), s(y'''), z'''))))
F(s(s(x'''')), 0, s(s(s(s(0))))) -> F(s(x''''), s(0), s(s(s(0))))


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(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(f(x1, x2, x3))=  0  
  POL(F(x1, x2, x3))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 33
Dependency Graph


Dependency Pairs:

F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(s(x''''')), s(s(y''''')), s(s(s(0)))) -> F(s(s(x''''')), s(s(y''''')), s(s(0)))
F(s(s(x''''')), s(0), s(s(s(0)))) -> F(s(s(x''''')), s(0), s(s(0)))
F(s(s(0)), s(y''''), s(s(s(0)))) -> F(s(s(0)), s(y''''), s(s(0)))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x'''), s(y'''), s(s(s(0)))) -> F(s(x'''), s(y'''), s(s(0)))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(s(0))) -> F(s(s(s(x'''''''))), s(s(s(y'''''''))), s(0))
F(s(s(s(x'''''''))), s(s(0)), s(s(0))) -> F(s(s(s(x'''''''))), s(s(0)), s(0))
F(s(s(s(0))), s(s(y''''''''')), s(s(0))) -> F(s(s(s(0))), s(s(y''''''''')), s(0))
F(s(s(x'''''')), s(s(y'''''')), s(s(0))) -> F(s(s(x'''''')), s(s(y'''''')), s(0))
F(s(s(x'''')), s(s(y'''')), s(s(0))) -> F(s(s(x'''')), s(s(y'''')), s(0))
F(s(s(x'''')), s(0), s(s(0))) -> F(s(s(x'''')), s(0), s(0))
F(s(s(0)), s(y'''), s(s(0))) -> F(s(s(0)), s(y'''), s(0))
F(s(x'), s(y'), s(s(s(0)))) -> F(s(x'), s(y'), s(s(0)))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(s(x''')), s(s(y''')), s(s(0))) -> F(s(s(x''')), s(s(y''')), s(0))
F(s(s(x''')), s(0), s(s(0))) -> F(s(s(x''')), s(0), s(0))
F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(s(y''''''''')))), 0)
F(s(s(s(s(x''''''''')))), s(s(s(0))), s(0)) -> F(s(s(s(s(x''''''''')))), s(s(s(0))), 0)
F(s(s(s(s(0)))), s(s(s(y'''''''''''))), s(0)) -> F(s(s(s(s(0)))), s(s(s(y'''''''''''))), 0)
F(s(s(s(x''''''''))), s(s(s(y''''''''))), s(0)) -> F(s(s(s(x''''''''))), s(s(s(y''''''''))), 0)
F(s(s(s(0))), s(s(y'''''')), s(0)) -> F(s(s(s(0))), s(s(y'''''')), 0)
F(s(s(s(x''''''))), s(s(s(y''''''))), s(0)) -> F(s(s(s(x''''''))), s(s(s(y''''''))), 0)
F(s(s(s(x''''''))), s(s(0)), s(0)) -> F(s(s(s(x''''''))), s(s(0)), 0)
F(s(s(s(0))), s(s(y''''')), s(0)) -> F(s(s(s(0))), s(s(y''''')), 0)
F(s(s(s(x'''''))), s(s(s(y'''''))), s(0)) -> F(s(s(s(x'''''))), s(s(s(y'''''))), 0)
F(s(s(s(x'''''))), s(s(0)), s(0)) -> F(s(s(s(x'''''))), s(s(0)), 0)
F(s(s(s(0))), s(s(y''''''')), s(0)) -> F(s(s(s(0))), s(s(y''''''')), 0)
F(s(s(0)), s(y'), s(s(0))) -> F(s(s(0)), s(y'), s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 34
Forward Instantiation Transformation


Dependency Pairs:

F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




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

F(s(s(0)), s(y'), s(s(s(z''''')))) -> F(s(s(0)), s(y'), s(s(z''''')))
seven new Dependency Pairs are created:

F(s(s(0)), s(y'''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y'''), s(s(s(z'''''''))))
F(s(s(0)), s(y'''), s(s(s(s(s(z'''''')))))) -> F(s(s(0)), s(y'''), s(s(s(s(z'''''')))))
F(s(s(0)), s(y''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''), s(s(s(z'''''''))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''')))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''''')))))
F(s(s(0)), s(y''), s(s(s(s(s(0)))))) -> F(s(s(0)), s(y''), s(s(s(s(0)))))
F(s(s(0)), s(y''), s(s(s(s(s(s(z'''''''))))))) -> F(s(s(0)), s(y''), s(s(s(s(s(z'''''''))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 35
Polynomial Ordering


Dependency Pairs:

F(s(s(0)), s(y''), s(s(s(s(s(s(z'''''''))))))) -> F(s(s(0)), s(y''), s(s(s(s(s(z'''''''))))))
F(s(s(0)), s(y''), s(s(s(s(s(0)))))) -> F(s(s(0)), s(y''), s(s(s(s(0)))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''''')))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''')))))
F(s(s(0)), s(y''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''), s(s(s(z'''''''))))
F(s(s(0)), s(y'''), s(s(s(s(s(z'''''')))))) -> F(s(s(0)), s(y'''), s(s(s(s(z'''''')))))
F(s(s(0)), s(y'''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y'''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

F(s(s(0)), s(y''), s(s(s(s(s(s(z'''''''))))))) -> F(s(s(0)), s(y''), s(s(s(s(s(z'''''''))))))
F(s(s(0)), s(y''), s(s(s(s(s(0)))))) -> F(s(s(0)), s(y''), s(s(s(s(0)))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''''')))))
F(s(s(0)), s(y''), s(s(s(s(s(z''''''')))))) -> F(s(s(0)), s(y''), s(s(s(s(z''''''')))))
F(s(s(0)), s(y''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''), s(s(s(z'''''''))))
F(s(s(0)), s(y'''), s(s(s(s(s(z'''''')))))) -> F(s(s(0)), s(y'''), s(s(s(s(z'''''')))))
F(s(s(0)), s(y'''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y'''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(s(s(z''''')))))) -> F(s(x''''), s(y''''), s(s(s(s(z''''')))))
F(s(s(0)), s(y''''), s(s(s(s(z'''''''))))) -> F(s(s(0)), s(y''''), s(s(s(z'''''''))))
F(s(x''''), s(y''''), s(s(s(z'''')))) -> F(s(x''''), s(y''''), s(s(z'''')))
F(s(x'''), s(y'''), s(s(s(s(z''''))))) -> F(s(x'''), s(y'''), s(s(s(z''''))))
F(s(x'), s(y'), s(s(s(s(z'''))))) -> F(s(x'), s(y'), s(s(s(z'''))))
F(s(s(0)), s(y'''), s(s(s(z''''')))) -> F(s(s(0)), s(y'''), s(s(z''''')))
F(s(x''''), s(y''''), s(s(s(s(0))))) -> F(s(x''''), s(y''''), s(s(s(0))))


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(0)=  0  
  POL(s(x1))=  1 + x1  
  POL(F(x1, x2, x3))=  1 + x1 + x2 + x3  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 36
Dependency Graph


Dependency Pair:


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:44 minutes