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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


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




The following dependency pair can be strictly oriented:

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


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(+'(x1, x2))=  x1 + x2  

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering


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




The following dependency pairs can be strictly oriented:

FIB(s(s(x))) -> FIB(x)
FIB(s(s(x))) -> FIB(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.
Used Argument Filtering System:
FIB(x1) -> FIB(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳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:
0:00 minutes