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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 4
Size-Change Principle
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. MINUS(s(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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
Size-Change Principle
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LE(s(x), s(y)) -> LE(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
UsableRules
       →DP Problem 3
Usable Rules (Innermost)


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Rewriting Transformation


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Rw
             ...
               →DP Problem 7
Negative Polynomial Order


Dependency Pair:

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


Rules:


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


Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Rw
             ...
               →DP Problem 8
Dependency Graph


Dependency Pair:


Rules:


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