Term Rewriting System R:
[y, x, z]
div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DIV(x, y) -> QUOT(x, y, y)
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> DIV(x, s(z))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
DIV(x, y) -> QUOT(x, y, y)


Rules:


div(0, y) -> 0
div(x, y) -> quot(x, y, y)
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(z)))


Strategy:

innermost




As we are in the innermost case, we can delete all 5 non-usable-rules.


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


Dependency Pairs:

QUOT(x, 0, s(z)) -> DIV(x, s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
DIV(x, y) -> QUOT(x, y, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. QUOT(x, 0, s(z)) -> DIV(x, s(z))
  2. QUOT(s(x), s(y), z) -> QUOT(x, y, z)
  3. DIV(x, y) -> QUOT(x, y, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
3=2
{2} , {2}
1>1
2>2
3=3
{3} , {3}
1=1
2=2
2=3

which lead(s) to this/these maximal multigraph(s):
{2} , {2}
1>1
2>2
3=3
{3} , {1}
1>1
2=2
{2} , {3}
1>1
3=2
3=3
{1} , {2}
1>1
3>2
3=3
{2} , {2}
1>1
3>2
3=3

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

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