Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx3_2_Luc97

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:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(dbl₁(s₁(X))) -> DBL₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(dbls₁(cons₂(X, Y))) -> DBLS₁(Y)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(dbls₁(X)) -> DBLS₁(proper₁(X))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
ACTIVE₁(indx₂(X1, X2)) -> INDX₂(active₁(X1), X2)
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> INDX₂(Y, Z)
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(dbls₁(X)) -> DBLS₁(active₁(X))
ACTIVE₁(dbls₁(X)) -> ACTIVE₁(X)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(dbls₁(cons₂(X, Y))) -> DBL₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
DBL₁(ok₁(X)) -> DBL₁(X)
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
DBLS₁(mark₁(X)) -> DBLS₁(X)
ACTIVE₁(dbl₁(X)) -> DBL₁(active₁(X))
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> CONS₂(sel₂(X, Z), indx₂(Y, Z))
PROPER₁(dbls₁(X)) -> PROPER₁(X)
PROPER₁(indx₂(X1, X2)) -> INDX₂(proper₁(X1), proper₁(X2))
S₁(ok₁(X)) -> S₁(X)
INDX₂(mark₁(X1), X2) -> INDX₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)
DBLS₁(ok₁(X)) -> DBLS₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
DBL₁(mark₁(X)) -> DBL₁(X)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(dbl₁(s₁(X))) -> S₁(dbl₁(X))
ACTIVE₁(dbls₁(cons₂(X, Y))) -> CONS₂(dbl₁(X), dbls₁(Y))
ACTIVE₁(from₁(X)) -> S₁(X)
PROPER₁(dbl₁(X)) -> DBL₁(proper₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
FROM₁(ok₁(X)) -> FROM₁(X)
INDX₂(ok₁(X1), ok₁(X2)) -> INDX₂(X1, X2)
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> SEL₂(X, Z)
ACTIVE₁(dbl₁(s₁(X))) -> S₁(s₁(dbl₁(X)))
PROPER₁(dbl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(dbl₁(s₁(X))) -> DBL₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(dbls₁(cons₂(X, Y))) -> DBLS₁(Y)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(dbls₁(X)) -> DBLS₁(proper₁(X))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
ACTIVE₁(indx₂(X1, X2)) -> INDX₂(active₁(X1), X2)
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> INDX₂(Y, Z)
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(dbls₁(X)) -> DBLS₁(active₁(X))
ACTIVE₁(dbls₁(X)) -> ACTIVE₁(X)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(dbls₁(cons₂(X, Y))) -> DBL₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
DBL₁(ok₁(X)) -> DBL₁(X)
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
DBLS₁(mark₁(X)) -> DBLS₁(X)
ACTIVE₁(dbl₁(X)) -> DBL₁(active₁(X))
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> CONS₂(sel₂(X, Z), indx₂(Y, Z))
PROPER₁(dbls₁(X)) -> PROPER₁(X)
PROPER₁(indx₂(X1, X2)) -> INDX₂(proper₁(X1), proper₁(X2))
S₁(ok₁(X)) -> S₁(X)
INDX₂(mark₁(X1), X2) -> INDX₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)
DBLS₁(ok₁(X)) -> DBLS₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
DBL₁(mark₁(X)) -> DBL₁(X)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(dbl₁(s₁(X))) -> S₁(dbl₁(X))
ACTIVE₁(dbls₁(cons₂(X, Y))) -> CONS₂(dbl₁(X), dbls₁(Y))
ACTIVE₁(from₁(X)) -> S₁(X)
PROPER₁(dbl₁(X)) -> DBL₁(proper₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
FROM₁(ok₁(X)) -> FROM₁(X)
INDX₂(ok₁(X1), ok₁(X2)) -> INDX₂(X1, X2)
ACTIVE₁(indx₂(cons₂(X, Y), Z)) -> SEL₂(X, Z)
ACTIVE₁(dbl₁(s₁(X))) -> S₁(s₁(dbl₁(X)))
PROPER₁(dbl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 10 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

FROM₁(ok₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INDX₂(ok₁(X1), ok₁(X2)) -> INDX₂(X1, X2)
INDX₂(mark₁(X1), X2) -> INDX₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

INDX₂(ok₁(X1), ok₁(X2)) -> INDX₂(X1, X2)

Used argument filtering: INDX₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INDX₂(mark₁(X1), X2) -> INDX₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

INDX₂(mark₁(X1), X2) -> INDX₂(X1, X2)

Used argument filtering: INDX₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBLS₁(ok₁(X)) -> DBLS₁(X)
DBLS₁(mark₁(X)) -> DBLS₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

DBLS₁(mark₁(X)) -> DBLS₁(X)

Used argument filtering: DBLS₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBLS₁(ok₁(X)) -> DBLS₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

DBLS₁(ok₁(X)) -> DBLS₁(X)

Used argument filtering: DBLS₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBL₁(mark₁(X)) -> DBL₁(X)
DBL₁(ok₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

DBL₁(ok₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBL₁(mark₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

DBL₁(mark₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(dbls₁(X)) -> PROPER₁(X)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(dbl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(indx₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
sel₂(x1, x2) = sel₂(x1, x2)
cons₂(x1, x2) = cons₂(x1, x2)
dbls₁(x1) = x1
indx₂(x1, x2) = indx₂(x1, x2)
dbl₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(dbls₁(X)) -> PROPER₁(X)
PROPER₁(dbl₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(dbl₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
dbls₁(x1) = x1
dbl₁(x1) = dbl₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(dbls₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(dbls₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
dbls₁(x1) = dbls₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(from₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(dbls₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
sel₂(x1, x2) = sel₂(x1, x2)
indx₂(x1, x2) = x1
dbl₁(x1) = x1
dbls₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)
ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(dbls₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ACTIVE₁(dbls₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = x1
indx₂(x1, x2) = x1
dbls₁(x1) = dbls₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
indx₂(x1, x2) = x1
dbl₁(x1) = dbl₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

ACTIVE₁(indx₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
indx₂(x1, x2) = indx₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(dbl₁(0)) -> mark₁(0)
active₁(dbl₁(s₁(X))) -> mark₁(s₁(s₁(dbl₁(X))))
active₁(dbls₁(nil)) -> mark₁(nil)
active₁(dbls₁(cons₂(X, Y))) -> mark₁(cons₂(dbl₁(X), dbls₁(Y)))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(indx₂(nil, X)) -> mark₁(nil)
active₁(indx₂(cons₂(X, Y), Z)) -> mark₁(cons₂(sel₂(X, Z), indx₂(Y, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(dbl₁(X)) -> dbl₁(active₁(X))
active₁(dbls₁(X)) -> dbls₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
active₁(indx₂(X1, X2)) -> indx₂(active₁(X1), X2)
dbl₁(mark₁(X)) -> mark₁(dbl₁(X))
dbls₁(mark₁(X)) -> mark₁(dbls₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
indx₂(mark₁(X1), X2) -> mark₁(indx₂(X1, X2))
proper₁(dbl₁(X)) -> dbl₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(dbls₁(X)) -> dbls₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(indx₂(X1, X2)) -> indx₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
dbl₁(ok₁(X)) -> ok₁(dbl₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
dbls₁(ok₁(X)) -> ok₁(dbls₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
indx₂(ok₁(X1), ok₁(X2)) -> ok₁(indx₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.