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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> CONCAT2(u, v)
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> LESS_LEAVES2(concat2(u, v), concat2(w, z))
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
REVERSE1(add2(n, x)) -> APP2(reverse1(x), add2(n, nil))
REVERSE1(add2(n, x)) -> REVERSE1(x)
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> CONCAT2(w, z)
SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
SHUFFLE1(add2(n, x)) -> REVERSE1(x)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
CONCAT2(cons2(u, v), y) -> CONCAT2(v, y)
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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:
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> CONCAT2(u, v)
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> LESS_LEAVES2(concat2(u, v), concat2(w, z))
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
REVERSE1(add2(n, x)) -> APP2(reverse1(x), add2(n, nil))
REVERSE1(add2(n, x)) -> REVERSE1(x)
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> CONCAT2(w, z)
SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
SHUFFLE1(add2(n, x)) -> REVERSE1(x)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
CONCAT2(cons2(u, v), y) -> CONCAT2(v, y)
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 7 SCCs with 5 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:
CONCAT2(cons2(u, v), y) -> CONCAT2(v, 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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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.
CONCAT2(cons2(u, v), y) -> CONCAT2(v, y)
Used argument filtering: CONCAT2(x1, x2) = x1
cons2(x1, x2) = cons1(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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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
Q DP problem:
The TRS P consists of the following rules:
LESS_LEAVES2(cons2(u, v), cons2(w, z)) -> LESS_LEAVES2(concat2(u, v), concat2(w, z))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
↳ 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:
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ 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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REVERSE1(add2(n, x)) -> REVERSE1(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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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.
REVERSE1(add2(n, x)) -> REVERSE1(x)
Used argument filtering: REVERSE1(x1) = x1
add2(x1, x2) = add1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ 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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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
Q DP problem:
The TRS P consists of the following rules:
SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
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)))
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))
concat2(leaf, y) -> y
concat2(cons2(u, v), y) -> cons2(u, concat2(v, y))
less_leaves2(x, leaf) -> false
less_leaves2(leaf, cons2(w, z)) -> true
less_leaves2(cons2(u, v), cons2(w, z)) -> less_leaves2(concat2(u, v), concat2(w, z))
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))
app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))
concat2(leaf, x0)
concat2(cons2(x0, x1), x2)
less_leaves2(x0, leaf)
less_leaves2(leaf, cons2(x0, x1))
less_leaves2(cons2(x0, x1), cons2(x2, x3))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.