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
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules


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




As we are in the innermost case, we can delete all 5 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. +'(X, s(Y)) -> +'(X, Y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Narrowing Transformation


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

F(0, s(0), X) -> F(X, +(X, X), X)
two new Dependency Pairs are created:

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

The transformation is resulting in no new DP problems.


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