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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
QUOTE1(sel2(X, Y)) -> SEL12(X, Y)
DBL11(s1(X)) -> DBL11(X)
QUOTE1(s1(X)) -> QUOTE1(X)
INDX2(cons2(X, Y), Z) -> SEL2(X, Z)
QUOTE1(dbl1(X)) -> DBL11(X)
DBLS1(cons2(X, Y)) -> DBL1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
DBLS1(cons2(X, Y)) -> DBLS1(Y)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
DBL1(s1(X)) -> DBL1(X)
FROM1(X) -> FROM1(s1(X))
INDX2(cons2(X, Y), Z) -> INDX2(Y, Z)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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, Y)) -> SEL12(X, Y)
DBL11(s1(X)) -> DBL11(X)
QUOTE1(s1(X)) -> QUOTE1(X)
INDX2(cons2(X, Y), Z) -> SEL2(X, Z)
QUOTE1(dbl1(X)) -> DBL11(X)
DBLS1(cons2(X, Y)) -> DBL1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
DBLS1(cons2(X, Y)) -> DBLS1(Y)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
DBL1(s1(X)) -> DBL1(X)
FROM1(X) -> FROM1(s1(X))
INDX2(cons2(X, Y), Z) -> INDX2(Y, Z)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 8 SCCs with 4 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:
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
Used argument filtering: SEL12(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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:
DBL11(s1(X)) -> DBL11(X)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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.
DBL11(s1(X)) -> DBL11(X)
Used argument filtering: DBL11(x1) = x1
s1(x1) = s1(x1)
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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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(s1(X)) -> QUOTE1(X)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ 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:
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INDX2(cons2(X, Y), Z) -> INDX2(Y, Z)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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.
INDX2(cons2(X, Y), Z) -> INDX2(Y, Z)
Used argument filtering: INDX2(x1, x2) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DBL1(s1(X)) -> DBL1(X)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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.
DBL1(s1(X)) -> DBL1(X)
Used argument filtering: DBL1(x1) = x1
s1(x1) = s1(x1)
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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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:
DBLS1(cons2(X, Y)) -> DBLS1(Y)
The TRS R consists of the following rules:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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.
DBLS1(cons2(X, Y)) -> DBLS1(Y)
Used argument filtering: DBLS1(x1) = x1
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:
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
dbls1(nil) -> nil
dbls1(cons2(X, Y)) -> cons2(dbl1(X), dbls1(Y))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
indx2(nil, X) -> nil
indx2(cons2(X, Y), Z) -> cons2(sel2(X, Z), indx2(Y, Z))
from1(X) -> cons2(X, from1(s1(X)))
dbl11(0) -> 01
dbl11(s1(X)) -> s11(s11(dbl11(X)))
sel12(0, cons2(X, Y)) -> X
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
quote1(0) -> 01
quote1(s1(X)) -> s11(quote1(X))
quote1(dbl1(X)) -> dbl11(X)
quote1(sel2(X, Y)) -> sel12(X, Y)
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.