Term Rewriting System R:
[X, Y, Z]
plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(plus(X, Y), Z) -> PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) -> PLUS(Y, Z)
TIMES(X, s(Y)) -> PLUS(X, times(Y, X))
TIMES(X, s(Y)) -> TIMES(Y, X)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)
one new Dependency Pair is created:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')


Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(plus(x1, x2))=  1 + x1 + x2  
  POL(PLUS(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

TIMES(X, s(Y)) -> TIMES(Y, X)


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

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

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳Forward Instantiation Transformation


Dependency Pair:

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




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

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))
one new Dependency Pair is created:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pair:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




The following dependency pair can be strictly oriented:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), 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(TIMES(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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