Term Rewriting System R:
[x, y]
f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, 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:

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

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Rw


Dependency Pair:

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


Rules:


f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, 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 w.r.t. to the AFS 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
         ↳Rw


Dependency Pair:


Rules:


f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, 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
         ↳Rewriting Transformation


Dependency Pairs:

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


Rules:


f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Rw
           →DP Problem 4
             ↳Argument Filtering and Ordering


Dependency Pairs:

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


Rules:


f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Rw
           →DP Problem 4
             ↳AFS
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, 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