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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE11(first2(X, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
FROM1(X) -> FROM1(s1(X))
QUOTE11(cons2(X, Z)) -> QUOTE1(X)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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:
QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE11(first2(X, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
FROM1(X) -> FROM1(s1(X))
QUOTE11(cons2(X, Z)) -> QUOTE1(X)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 8 SCCs with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
Used argument filtering: QUOTE1(x1) = x1
sel2(x1, x2) = x2
SEL12(x1, x2) = x2
s1(x1) = x1
cons2(x1, x2) = cons2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE1(s1(X)) -> QUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOTE1(s1(X)) -> QUOTE1(X)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.
QUOTE1(s1(X)) -> QUOTE1(X)
Used argument filtering: QUOTE1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
Used argument filtering: FIRST12(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
↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
Used argument filtering: QUOTE11(x1) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ 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:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
Used argument filtering: FIRST2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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(X), cons2(Y, Z)) -> SEL2(X, Z)
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
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)
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.