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
Argument Filtering and Ordering
       →DP Problem 2
Nar


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




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> x2
app(x1, x2) -> app(x1, x2)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 3
Dependency Graph
       →DP Problem 2
Nar


Dependency Pair:


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




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Narrowing Transformation


Dependency Pairs:

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




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

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Narrowing 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(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




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

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
two new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Nar
             ...
               →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 4
Nar
             ...
               →DP Problem 6
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes