Term Rewriting System R:
[x, l, y]
app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
APP(rev, app(app(cons, x), l)) -> APP(cons, app(app(rev1, x), l))
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev1, x)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev2, x)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(rev1, y)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev2, y)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
five new Dependency Pairs are created:

APP(rev, app(app(cons, 0), nil)) -> APP(app(cons, 0), app(app(rev2, 0), nil))
APP(rev, app(app(cons, app(s, x'')), nil)) -> APP(app(cons, app(s, x'')), app(app(rev2, app(s, x'')), nil))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(rev2, x''), app(app(cons, y'), l'')))
APP(rev, app(app(cons, x''), nil)) -> APP(app(cons, app(app(rev1, x''), nil)), nil)
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, x''), app(app(cons, y'), l''))), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rewriting Transformation


Dependency Pairs:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, x''), app(app(cons, y'), l''))), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(rev2, x''), app(app(cons, y'), l'')))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(rev2, x''), app(app(cons, y'), l'')))
one new Dependency Pair is created:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 3
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, x''), app(app(cons, y'), l''))), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, x''), app(app(cons, y'), l''))), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
one new Dependency Pair is created:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 4
                 ↳Rewriting Transformation


Dependency Pairs:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
one new Dependency Pair is created:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 5
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(rev, app(app(cons, x''), app(app(rev2, y'), l''))))
one new Dependency Pair is created:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 6
                 ↳Narrowing Transformation


Dependency Pairs:

APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
two new Dependency Pairs are created:

APP(app(rev2, x), app(app(cons, y'), nil)) -> APP(app(cons, x), nil)
APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(rev, app(app(cons, y0), app(app(rev2, y''), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 7
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(rev, app(app(cons, y0), app(app(rev2, y''), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




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

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(rev, app(app(cons, y0), app(app(rev2, y''), l''))))
one new Dependency Pair is created:

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(app(cons, app(app(rev1, y0), app(app(rev2, y''), l''))), app(app(rev2, y0), app(app(rev2, y''), l''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 8
                 ↳Argument Filtering and Ordering


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(app(cons, app(app(rev1, y0), app(app(rev2, y''), l''))), app(app(rev2, y0), app(app(rev2, y''), l''))))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)


The following usable rules for innermost can be oriented:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{cons, rev2, rev} > nil
{cons, rev2, rev} > {rev1, 0} > s

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> x1
app(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 9
                 ↳Argument Filtering and Ordering


Dependency Pairs:

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(app(cons, app(app(rev1, y0), app(app(rev2, y''), l''))), app(app(rev2, y0), app(app(rev2, y''), l''))))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

APP(app(rev2, x), app(app(cons, y0), app(app(cons, y''), l''))) -> APP(app(cons, x), app(app(cons, app(app(rev1, y0), app(app(rev2, y''), l''))), app(app(rev2, y0), app(app(rev2, y''), l''))))
APP(rev, app(app(cons, x''), app(app(cons, y'), l''))) -> APP(app(cons, app(app(rev1, y'), l'')), app(app(cons, app(app(rev1, x''), app(app(rev2, y'), l''))), app(app(rev2, x''), app(app(rev2, y'), l''))))


The following usable rules for innermost can be oriented:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{rev1, 0} > s
{rev2, rev} > nil
{rev2, rev} > cons

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> x1
app(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Rw
             ...
               →DP Problem 10
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)


Rules:


app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:16 minutes