Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx2_Luc03b

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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(from₁(X)) -> FROM₁(s₁(X))
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(fst₂(X1, X2)) -> FST₂(active₁(X1), X2)
PROPER₁(len₁(X)) -> LEN₁(proper₁(X))
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(fst₂(s₁(X), cons₂(Y, Z))) -> FST₂(X, Z)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(len₁(X)) -> PROPER₁(X)
ACTIVE₁(fst₂(s₁(X), cons₂(Y, Z))) -> CONS₂(Y, fst₂(X, Z))
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
ACTIVE₁(len₁(X)) -> LEN₁(active₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
LEN₁(ok₁(X)) -> LEN₁(X)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(X1, active₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(fst₂(X1, X2)) -> FST₂(X1, active₁(X2))
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
FST₂(X1, mark₁(X2)) -> FST₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(fst₂(X1, X2)) -> FST₂(proper₁(X1), proper₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(len₁(cons₂(X, Z))) -> S₁(len₁(Z))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
LEN₁(mark₁(X)) -> LEN₁(X)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(len₁(cons₂(X, Z))) -> LEN₁(Z)
FST₂(ok₁(X1), ok₁(X2)) -> FST₂(X1, X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(len₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> S₁(X)
ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
FST₂(mark₁(X1), X2) -> FST₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(from₁(X)) -> FROM₁(s₁(X))
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(fst₂(X1, X2)) -> FST₂(active₁(X1), X2)
PROPER₁(len₁(X)) -> LEN₁(proper₁(X))
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(fst₂(s₁(X), cons₂(Y, Z))) -> FST₂(X, Z)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(len₁(X)) -> PROPER₁(X)
ACTIVE₁(fst₂(s₁(X), cons₂(Y, Z))) -> CONS₂(Y, fst₂(X, Z))
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
ACTIVE₁(len₁(X)) -> LEN₁(active₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
LEN₁(ok₁(X)) -> LEN₁(X)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(X1, active₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(fst₂(X1, X2)) -> FST₂(X1, active₁(X2))
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
FST₂(X1, mark₁(X2)) -> FST₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(fst₂(X1, X2)) -> FST₂(proper₁(X1), proper₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(len₁(cons₂(X, Z))) -> S₁(len₁(Z))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
LEN₁(mark₁(X)) -> LEN₁(X)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(len₁(cons₂(X, Z))) -> LEN₁(Z)
FST₂(ok₁(X1), ok₁(X2)) -> FST₂(X1, X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(len₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> S₁(X)
ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
FST₂(mark₁(X1), X2) -> FST₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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 9 SCCs with 24 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

            ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

LEN₁(mark₁(X)) -> LEN₁(X)
LEN₁(ok₁(X)) -> LEN₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

LEN₁(ok₁(X)) -> LEN₁(X)

Used argument filtering: LEN₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEN₁(mark₁(X)) -> LEN₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

LEN₁(mark₁(X)) -> LEN₁(X)

Used argument filtering: LEN₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

FROM₁(mark₁(X)) -> FROM₁(X)
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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
mark₁(x1) = x1
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

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

FROM₁(mark₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(mark₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = 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

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

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

FST₂(ok₁(X1), ok₁(X2)) -> FST₂(X1, X2)
FST₂(mark₁(X1), X2) -> FST₂(X1, X2)
FST₂(X1, mark₁(X2)) -> FST₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

FST₂(X1, mark₁(X2)) -> FST₂(X1, X2)

Used argument filtering: FST₂(x1, x2) = x2
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FST₂(ok₁(X1), ok₁(X2)) -> FST₂(X1, X2)
FST₂(mark₁(X1), X2) -> FST₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

FST₂(ok₁(X1), ok₁(X2)) -> FST₂(X1, X2)

Used argument filtering: FST₂(x1, x2) = x2
ok₁(x1) = ok₁(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

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

FST₂(mark₁(X1), X2) -> FST₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.

FST₂(mark₁(X1), X2) -> FST₂(X1, X2)

Used argument filtering: FST₂(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

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

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = 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

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(len₁(X)) -> PROPER₁(X)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(fst₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
add₂(x1, x2) = add₂(x1, x2)
fst₂(x1, x2) = fst₂(x1, x2)
s₁(x1) = x1
from₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
len₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(len₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(len₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
len₁(x1) = len₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

            ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

            ↳ 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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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

ACTIVE₁(len₁(X)) -> ACTIVE₁(X)
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(fst₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(fst₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)

Used argument filtering: ACTIVE₁(x1) = x1
len₁(x1) = x1
fst₂(x1, x2) = fst₂(x1, x2)
cons₂(x1, x2) = x1
add₂(x1, x2) = add₂(x1, x2)
from₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(len₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(from₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
len₁(x1) = x1
cons₂(x1, x2) = 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

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

ACTIVE₁(len₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
len₁(x1) = x1
cons₂(x1, x2) = cons₁(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

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

ACTIVE₁(len₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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(len₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
len₁(x1) = len₁(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

                            ↳ PisEmptyProof

          ↳ QDP

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

active₁(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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

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(fst₂(0, Z)) -> mark₁(nil)
active₁(fst₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, fst₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(len₁(nil)) -> mark₁(0)
active₁(len₁(cons₂(X, Z))) -> mark₁(s₁(len₁(Z)))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(active₁(X1), X2)
active₁(fst₂(X1, X2)) -> fst₂(X1, active₁(X2))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
active₁(len₁(X)) -> len₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
fst₂(mark₁(X1), X2) -> mark₁(fst₂(X1, X2))
fst₂(X1, mark₁(X2)) -> mark₁(fst₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
len₁(mark₁(X)) -> mark₁(len₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(fst₂(X1, X2)) -> fst₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(len₁(X)) -> len₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
fst₂(ok₁(X1), ok₁(X2)) -> ok₁(fst₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
len₁(ok₁(X)) -> ok₁(len₁(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.