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
         ↳Forward Instantiation Transformation
       →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




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
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Rw


Dependency Pair:

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




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

+'(X'', s(s(Y''))) -> +'(X'', s(Y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Rw


Dependency Pair:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳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
         ↳FwdInst
       →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
         ↳FwdInst
       →DP Problem 2
         ↳Rw
           →DP Problem 6
             ↳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