Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
QUICKSORT1(add2(n, x)) -> HIGH2(n, x)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
LOW2(n, add2(m, x)) -> LE2(m, n)
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
LE2(s1(x), s1(y)) -> LE2(x, y)
QUICKSORT1(add2(n, x)) -> LOW2(n, x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
QUICKSORT1(add2(n, x)) -> HIGH2(n, x)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
LOW2(n, add2(m, x)) -> LE2(m, n)
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
LE2(s1(x), s1(y)) -> LE2(x, y)
QUICKSORT1(add2(n, x)) -> LOW2(n, x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 7 SCCs with 6 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(add2(n, x), y) -> APP2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(add2(n, x), y) -> APP2(x, y)
Used argument filtering: APP2(x1, x2) = x1
add2(x1, x2) = add1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE2(s1(x), s1(y)) -> LE2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
Used argument filtering: IF_HIGH3(x1, x2, x3) = x3
add2(x1, x2) = add1(x2)
HIGH2(x1, x2) = x2
le2(x1, x2) = le
0 = 0
true = true
s1(x1) = s
false = false
Used ordering: Precedence:
le > true
le > false
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
Used argument filtering: LOW2(x1, x2) = x2
add2(x1, x2) = add1(x2)
IF_LOW3(x1, x2, x3) = x3
le2(x1, x2) = le
0 = 0
true = true
s1(x1) = s
false = false
Used ordering: Precedence:
le > true
le > false
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
minus2(x1, x2) = x1
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
The set Q consists of the following terms:
minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.