Term Rewriting System R:
[X, Y]
+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(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)
DOUBLE(X) -> +'(X, X)
F(0, s(0), X) -> F(X, double(X), X)
F(0, s(0), X) -> DOUBLE(X)

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:


+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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:


+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(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
         ↳AFS
       →DP Problem 2
         ↳Rewriting Transformation


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

F(0, s(0), X) -> F(X, double(X), X)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Rw
           →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))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> 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