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
         ↳Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 2 non-usable-rules.


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


Dependency Pair:

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


Rules:


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
         ↳UsableRules
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 3
                 ↳Rewriting Transformation


Dependency Pair:

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


Rules:


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
         ↳UsableRules
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 4
                 ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 2 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 5
                 ↳Size-Change Principle


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. TIMES(x, plus(y, 1)) -> TIMES(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:
plus(x1, x2) -> plus(x1, x2)

We obtain no new DP problems.

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