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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(minus, x), app(s, y)) -> APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(minus, x)

Furthermore, R contains two SCCs.


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


Dependency Pair:

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


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(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 3
A-Transformation
       →DP Problem 2
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 3
ATrans
             ...
               →DP Problem 4
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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)


Dependency Pair:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))


Rules:


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


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
UsableRules
           →DP Problem 5
A-Transformation


Dependency Pair:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))


Rules:


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


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
ATrans
             ...
               →DP Problem 6
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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
ATrans
             ...
               →DP Problem 7
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.

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