Term Rewriting System R:
[X, Y]
+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
f(0, s(0), X) -> f(X, +(X, X), X)
g(X, Y) -> X
g(X, Y) -> Y

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(X, s(Y)) -> +'(X, Y)
F(0, s(0), X) -> F(X, +(X, X), X)
F(0, s(0), X) -> +'(X, X)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pair:

+'(X, s(Y)) -> +'(X, Y)


Rules:


+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
f(0, s(0), X) -> f(X, +(X, X), X)
g(X, Y) -> X
g(X, Y) -> 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))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
f(0, s(0), X) -> f(X, +(X, X), X)
g(X, Y) -> X
g(X, Y) -> Y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(0, s(0), X) -> F(X, +(X, X), X)


Rules:


+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
f(0, s(0), X) -> f(X, +(X, X), X)
g(X, Y) -> X
g(X, Y) -> Y


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes