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:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
FROM1(X) -> FROM1(s1(X))
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ 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))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
FROM1(X) -> FROM1(s1(X))
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ 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:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
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: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
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:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
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) = s
minus2(x1, x2) = minus
0 = 0
Used ordering: Quasi Precedence:
s > minus > 0
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
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.
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
Used argument filtering: ZWQUOT2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
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.
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
Used argument filtering: SEL2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FROM1(X) -> FROM1(s1(X))
The TRS R consists of the following rules:
from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)
minus2(X, 0) -> 0
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)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), zWquot2(XS, YS))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.