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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) -> PLUS(z, s(0))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules


Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)


Rules:


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


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 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. PLUS(s(x), y) -> PLUS(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:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Negative Polynomial Order


Dependency Pairs:

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


Rules:


plus(s(x), y) -> s(plus(x, y))
plus(0, y) -> y


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

plus(s(x), y) -> s(plus(x, y))
plus(0, y) -> y


Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, ..., x3) ) = x1

POL( s(x1) ) = x1 + 1

POL( plus(x1, x2) ) = x1 + x2


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Neg POLO
             ...
               →DP Problem 5
                 ↳Narrowing Transformation


Dependency Pair:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))


Rules:


plus(s(x), y) -> s(plus(x, y))
plus(0, y) -> y


Strategy:

innermost




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

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))

The transformation is resulting in no new DP problems.


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