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:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, X2)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, X2)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__0) -> 01
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
ACTIVATE1(n__s1(X)) -> S1(X)
UNQUOTE1(01) -> 01
QUOTE1(n__sel2(X, Z)) -> SEL12(activate1(X), activate1(Z))
QUOTE11(n__cons2(X, Z)) -> QUOTE11(activate1(Z))
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
QUOTE11(n__first2(X, Z)) -> FIRST12(activate1(X), activate1(Z))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(X)
FIRST2(0, Z) -> NIL
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(X1, X2)
UNQUOTE11(nil1) -> NIL
QUOTE11(n__first2(X, Z)) -> ACTIVATE1(X)
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(Z)
FCONS2(X, Z) -> CONS2(X, Z)
QUOTE11(n__cons2(X, Z)) -> QUOTE1(activate1(X))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
FROM1(X) -> CONS2(X, n__from1(s1(X)))
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(X)
FROM1(X) -> S1(X)
UNQUOTE1(s11(X)) -> S1(unquote1(X))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
SEL12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
FIRST12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, activate1(Z))
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
QUOTE1(n__s1(X)) -> QUOTE1(activate1(X))
QUOTE1(n__s1(X)) -> ACTIVATE1(X)
QUOTE11(n__first2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> CONS2(Y, n__first2(X, activate1(Z)))
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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:
ACTIVATE1(n__0) -> 01
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
ACTIVATE1(n__s1(X)) -> S1(X)
UNQUOTE1(01) -> 01
QUOTE1(n__sel2(X, Z)) -> SEL12(activate1(X), activate1(Z))
QUOTE11(n__cons2(X, Z)) -> QUOTE11(activate1(Z))
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
QUOTE11(n__first2(X, Z)) -> FIRST12(activate1(X), activate1(Z))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(X)
FIRST2(0, Z) -> NIL
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(X1, X2)
UNQUOTE11(nil1) -> NIL
QUOTE11(n__first2(X, Z)) -> ACTIVATE1(X)
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(Z)
FCONS2(X, Z) -> CONS2(X, Z)
QUOTE11(n__cons2(X, Z)) -> QUOTE1(activate1(X))
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
FROM1(X) -> CONS2(X, n__from1(s1(X)))
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(X)
FROM1(X) -> S1(X)
UNQUOTE1(s11(X)) -> S1(unquote1(X))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
SEL12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
FIRST12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, activate1(Z))
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
QUOTE1(n__s1(X)) -> QUOTE1(activate1(X))
QUOTE1(n__s1(X)) -> ACTIVATE1(X)
QUOTE11(n__first2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> CONS2(Y, n__first2(X, activate1(Z)))
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 6 SCCs with 27 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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.
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
Used argument filtering: UNQUOTE1(x1) = x1
s11(x1) = s11(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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.
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
Used argument filtering: UNQUOTE11(x1) = x1
cons12(x1, x2) = cons11(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(X1, X2)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, activate1(Z))
QUOTE1(n__sel2(X, Z)) -> SEL12(activate1(X), activate1(Z))
QUOTE1(n__s1(X)) -> QUOTE1(activate1(X))
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, 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
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOTE11(n__cons2(X, Z)) -> QUOTE11(activate1(Z))
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, activate1(Z))
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, activate1(Z)))
quote1(n__0) -> 01
quote11(n__cons2(X, Z)) -> cons12(quote1(activate1(X)), quote11(activate1(Z)))
quote11(n__nil) -> nil1
quote1(n__s1(X)) -> s11(quote1(activate1(X)))
quote1(n__sel2(X, Z)) -> sel12(activate1(X), activate1(Z))
quote11(n__first2(X, Z)) -> first12(activate1(X), activate1(Z))
unquote1(01) -> 0
unquote1(s11(X)) -> s1(unquote1(X))
unquote11(nil1) -> nil
unquote11(cons12(X, Z)) -> fcons2(unquote1(X), unquote11(Z))
fcons2(X, Z) -> cons2(X, Z)
first2(X1, X2) -> n__first2(X1, X2)
from1(X) -> n__from1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
s1(X) -> n__s1(X)
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(n__nil) -> nil
activate1(n__s1(X)) -> s1(X)
activate1(n__sel2(X1, X2)) -> sel2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.