Term Rewriting System R:
[x]
fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FIB(s(s(x))) -> FIB(s(x))
FIB(s(s(x))) -> FIB(x)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

FIB(s(s(x))) -> FIB(x)
FIB(s(s(x))) -> FIB(s(x))


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




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

FIB(s(s(x))) -> FIB(s(x))
one new Dependency Pair is created:

FIB(s(s(s(x'')))) -> FIB(s(s(x'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳Forward Instantiation Transformation


Dependency Pairs:

FIB(s(s(s(x'')))) -> FIB(s(s(x'')))
FIB(s(s(x))) -> FIB(x)


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




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

FIB(s(s(x))) -> FIB(x)
two new Dependency Pairs are created:

FIB(s(s(s(s(x''))))) -> FIB(s(s(x'')))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(x''''))))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(x''))))) -> FIB(s(s(x'')))
FIB(s(s(s(x'')))) -> FIB(s(s(x'')))


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




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

FIB(s(s(s(x'')))) -> FIB(s(s(x'')))
three new Dependency Pairs are created:

FIB(s(s(s(s(x''''))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(s(x''''''))))))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(x''''))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(x''))))) -> FIB(s(s(x'')))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(x''''))))


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




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

FIB(s(s(s(s(x''))))) -> FIB(s(s(x'')))
four new Dependency Pairs are created:

FIB(s(s(s(s(s(s(x''''))))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''))))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(x''''))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(s(x''''''))))))


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




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

FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(x''''))))
six new Dependency Pairs are created:

FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))
FIB(s(s(s(s(s(s(s(s(x''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''')))))))
FIB(s(s(s(s(s(s(s(s(s(x'''''''')))))))))) -> FIB(s(s(s(s(s(s(s(x''''''''))))))))
FIB(s(s(s(s(s(s(s(s(s(s(x''''''''''))))))))))) -> FIB(s(s(s(s(s(s(s(s(x'''''''''')))))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pairs:

FIB(s(s(s(s(s(s(s(s(s(s(x''''''''''))))))))))) -> FIB(s(s(s(s(s(s(s(s(x'''''''''')))))))))
FIB(s(s(s(s(s(s(s(s(s(x'''''''')))))))))) -> FIB(s(s(s(s(s(s(s(x''''''''))))))))
FIB(s(s(s(s(s(s(s(s(x''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''')))))))
FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''))))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(x''''))))) -> FIB(s(s(s(x''''))))
FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

FIB(s(s(s(s(s(s(s(s(s(s(x''''''''''))))))))))) -> FIB(s(s(s(s(s(s(s(s(x'''''''''')))))))))
FIB(s(s(s(s(s(s(s(s(s(x'''''''')))))))))) -> FIB(s(s(s(s(s(s(s(x''''''''))))))))
FIB(s(s(s(s(s(s(s(s(x''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''')))))))
FIB(s(s(s(s(s(s(s(s(x''''''''))))))))) -> FIB(s(s(s(s(s(s(x'''''''')))))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(x'''''')))))
FIB(s(s(s(s(s(s(s(x'''''')))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(s(x''''))))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(s(s(x''''''))))))) -> FIB(s(s(s(s(s(x''''''))))))
FIB(s(s(s(s(s(x'''')))))) -> FIB(s(s(s(s(x'''')))))
FIB(s(s(s(s(x''''))))) -> FIB(s(s(s(x''''))))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  
  POL(FIB(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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