Term Rewriting System R:
[y, x]
-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(x, s(y)) -> -'(x, p(s(y)))
-'(x, s(y)) -> P(s(y))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Rewriting Transformation


Dependency Pair:

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


Rules:


-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

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


Rules:


-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x


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
         ↳Rw
           →DP Problem 2
             ↳FwdInst
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(-'(x1, x2))=  1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes