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

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(x))
FIB(s(s(x))) -> FIB(s(x))
FIB(s(s(x))) -> FIB(x)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

+'(x, s(y)) -> +'(x, y)


Rules:


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


Strategy:

innermost




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

+'(x, s(y)) -> +'(x, y)
one new Dependency Pair is created:

+'(x'', s(s(y''))) -> +'(x'', s(y''))

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

+'(x'', s(s(y''))) -> +'(x'', s(y''))


Rules:


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


Strategy:

innermost




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

+'(x'', s(s(y''))) -> +'(x'', s(y''))
one new Dependency Pair is created:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳FwdInst


Dependency Pair:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(y'''')))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(y'''')))


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(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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
         ↳FwdInst
           →DP Problem 6
             ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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 6
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳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))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


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 w.r.t. to the implicit AFS 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 6
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes