Term Rewriting System R:
[x, y, z, k, l]
app'(app'(minus, x), 0) -> x
app'(app'(minus, app'(s, x)), app'(s, y)) -> app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) -> app'(app'(minus, x), app'(app'(plus, y), z))
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)))
app'(app'(plus, 0), y) -> y
app'(app'(plus, app'(s, x)), y) -> app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) -> k
app'(app'(app, l), nil) -> l
app'(app'(app, app'(app'(cons, x), l)), k) -> app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) -> app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

APP'(app'(minus, app'(s, x)), app'(s, y)) -> APP'(app'(minus, x), y)
APP'(app'(minus, app'(s, x)), app'(s, y)) -> APP'(minus, x)
APP'(app'(minus, app'(app'(minus, x), y)), z) -> APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(minus, app'(app'(minus, x), y)), z) -> APP'(app'(plus, y), z)
APP'(app'(minus, app'(app'(minus, x), y)), z) -> APP'(plus, 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)
APP'(app'(plus, app'(s, x)), y) -> APP'(s, app'(app'(plus, x), y))
APP'(app'(plus, app'(s, x)), y) -> APP'(app'(plus, x), y)
APP'(app'(plus, app'(s, x)), y) -> APP'(plus, x)
APP'(app'(app, app'(app'(cons, x), l)), k) -> APP'(app'(cons, x), app'(app'(app, l), k))
APP'(app'(app, app'(app'(cons, x), l)), k) -> APP'(app'(app, l), k)
APP'(app'(app, app'(app'(cons, x), l)), k) -> APP'(app, l)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(app'(cons, app'(app'(plus, x), y)), l)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(cons, app'(app'(plus, x), y))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(app'(plus, x), y)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(plus, x)
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> APP'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k))))
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> APP'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> APP'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))
APP'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> APP'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(app'(plus, x), y)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(app'(cons, app'(app'(plus, x), y)), l)
APP'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> APP'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
APP'(app'(app, app'(app'(cons, x), l)), k) -> APP'(app'(app, l), k)
APP'(app'(plus, app'(s, x)), y) -> APP'(app'(plus, 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'(app'(quot, app'(app'(minus, x), y)), app'(s, y))
APP'(app'(minus, app'(app'(minus, x), y)), z) -> APP'(app'(plus, y), z)
APP'(app'(minus, app'(app'(minus, x), y)), z) -> APP'(app'(minus, x), app'(app'(plus, y), z))
APP'(app'(minus, app'(s, x)), app'(s, y)) -> APP'(app'(minus, x), y)


Rules:


app'(app'(minus, x), 0) -> x
app'(app'(minus, app'(s, x)), app'(s, y)) -> app'(app'(minus, x), y)
app'(app'(minus, app'(app'(minus, x), y)), z) -> app'(app'(minus, x), app'(app'(plus, y), z))
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)))
app'(app'(plus, 0), y) -> y
app'(app'(plus, app'(s, x)), y) -> app'(s, app'(app'(plus, x), y))
app'(app'(app, nil), k) -> k
app'(app'(app, l), nil) -> l
app'(app'(app, app'(app'(cons, x), l)), k) -> app'(app'(cons, x), app'(app'(app, l), k))
app'(sum, app'(app'(cons, x), nil)) -> app'(app'(cons, x), nil)
app'(sum, app'(app'(cons, x), app'(app'(cons, y), l))) -> app'(sum, app'(app'(cons, app'(app'(plus, x), y)), l))
app'(sum, app'(app'(app, l), app'(app'(cons, x), app'(app'(cons, y), k)))) -> app'(sum, app'(app'(app, l), app'(sum, app'(app'(cons, x), app'(app'(cons, y), k)))))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes