Term Rewriting System R:
[x, y]
times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) -> x
times(x, 0) -> 0
plus(x, 0) -> x

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TIMES(x, plus(y, 1)) -> PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) -> PLUS(y, times(1, 0))
TIMES(x, plus(y, 1)) -> TIMES(1, 0)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Rewriting Transformation


Dependency Pair:

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))


Rules:


times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) -> x
times(x, 0) -> 0
plus(x, 0) -> x


Strategy:

innermost




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

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))
one new Dependency Pair is created:

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Rw
           →DP Problem 2
             ↳Rewriting Transformation


Dependency Pair:

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, 0))


Rules:


times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) -> x
times(x, 0) -> 0
plus(x, 0) -> x


Strategy:

innermost




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

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, 0))
one new Dependency Pair is created:

TIMES(x, plus(y, 1)) -> TIMES(x, y)

The transformation is resulting in one new DP problem: 



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


Dependency Pair:

TIMES(x, plus(y, 1)) -> TIMES(x, y)


Rules:


times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) -> x
times(x, 0) -> 0
plus(x, 0) -> x


Strategy:

innermost




The following dependency pair can be strictly oriented:

TIMES(x, plus(y, 1)) -> TIMES(x, 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))=  x2  
  POL(plus(x1, x2))=  1 + x1  
  POL(1)=  0  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) -> x
times(x, 0) -> 0
plus(x, 0) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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