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, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
QUOT2(s1(X), s1(Y)) -> S1(quot2(minus2(X, Y), s1(Y)))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
QUOT2(s1(X), s1(Y)) -> S1(quot2(minus2(X, Y), s1(Y)))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ 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, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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
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, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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
n__s1(x1) = n__s
Used ordering: Quasi Precedence:
s > minus > 0
s > n__s
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
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
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(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), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.