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

Termination of R to be shown. 



   R
     ↳Overlay and local confluence Check



The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost. 


   R
     ↳OC
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains two SCCs. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

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


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))


Strategy:

innermost




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


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳Size-Change Principle
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MINUS(x, s(y)) -> MINUS(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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))


Strategy:

innermost




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


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Negative Polynomial Order


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x


Strategy:

innermost




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

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


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

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x


Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( pred(x1) ) = x1


This results in one new DP problem. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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