Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx15_Luc98

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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

FIRST₂(X1, mark₁(X2)) -> FIRST₂(X1, X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
PROPER₁(and₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(and₂(X1, X2)) -> AND₂(proper₁(X1), proper₁(X2))
FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
ACTIVE₁(first₂(X1, X2)) -> FIRST₂(active₁(X1), X2)
ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
AND₂(mark₁(X1), X2) -> AND₂(X1, X2)
AND₂(ok₁(X1), ok₁(X2)) -> AND₂(X1, X2)
ACTIVE₁(first₂(s₁(X), cons₂(Y, Z))) -> FIRST₂(X, Z)
TOP₁(mark₁(X)) -> PROPER₁(X)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
ACTIVE₁(first₂(X1, X2)) -> FIRST₂(X1, active₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(first₂(s₁(X), cons₂(Y, Z))) -> CONS₂(Y, first₂(X, Z))
S₁(ok₁(X)) -> S₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
PROPER₁(first₂(X1, X2)) -> FIRST₂(proper₁(X1), proper₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> S₁(X)
FIRST₂(ok₁(X1), ok₁(X2)) -> FIRST₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
PROPER₁(and₂(X1, X2)) -> PROPER₁(X2)
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(and₂(X1, X2)) -> AND₂(active₁(X1), X2)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

FIRST₂(X1, mark₁(X2)) -> FIRST₂(X1, X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
PROPER₁(and₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(and₂(X1, X2)) -> AND₂(proper₁(X1), proper₁(X2))
FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
ACTIVE₁(first₂(X1, X2)) -> FIRST₂(active₁(X1), X2)
ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
AND₂(mark₁(X1), X2) -> AND₂(X1, X2)
AND₂(ok₁(X1), ok₁(X2)) -> AND₂(X1, X2)
ACTIVE₁(first₂(s₁(X), cons₂(Y, Z))) -> FIRST₂(X, Z)
TOP₁(mark₁(X)) -> PROPER₁(X)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X2)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
ACTIVE₁(first₂(X1, X2)) -> FIRST₂(X1, active₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
ACTIVE₁(first₂(s₁(X), cons₂(Y, Z))) -> CONS₂(Y, first₂(X, Z))
S₁(ok₁(X)) -> S₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
PROPER₁(first₂(X1, X2)) -> FIRST₂(proper₁(X1), proper₁(X2))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(from₁(X)) -> S₁(X)
FIRST₂(ok₁(X1), ok₁(X2)) -> FIRST₂(X1, X2)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
PROPER₁(and₂(X1, X2)) -> PROPER₁(X2)
FROM₁(ok₁(X)) -> FROM₁(X)
ACTIVE₁(and₂(X1, X2)) -> AND₂(active₁(X1), X2)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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 21 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

FIRST₂(X1, mark₁(X2)) -> FIRST₂(X1, X2)
FIRST₂(ok₁(X1), ok₁(X2)) -> FIRST₂(X1, X2)
FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

FIRST₂(ok₁(X1), ok₁(X2)) -> FIRST₂(X1, X2)

Used argument filtering: FIRST₂(x1, x2) = x2
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

          ↳ QDP

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

FIRST₂(X1, mark₁(X2)) -> FIRST₂(X1, X2)
FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

FIRST₂(X1, mark₁(X2)) -> FIRST₂(X1, X2)

Used argument filtering: FIRST₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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:

FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

FIRST₂(mark₁(X1), X2) -> FIRST₂(X1, X2)

Used argument filtering: FIRST₂(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

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

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

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

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

Used argument filtering: ADD₂(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

          ↳ 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

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

          ↳ 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = x3
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

          ↳ QDP

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

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = 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

          ↳ QDP

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

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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:

AND₂(ok₁(X1), ok₁(X2)) -> AND₂(X1, X2)
AND₂(mark₁(X1), X2) -> AND₂(X1, X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

AND₂(ok₁(X1), ok₁(X2)) -> AND₂(X1, X2)

Used argument filtering: AND₂(x1, x2) = x2
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:

AND₂(mark₁(X1), X2) -> AND₂(X1, X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.

AND₂(mark₁(X1), X2) -> AND₂(X1, X2)

Used argument filtering: AND₂(x1, x2) = 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(and₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(first₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(and₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
and₂(x1, x2) = and₂(x1, x2)
cons₂(x1, x2) = cons₂(x1, x2)
if₃(x1, x2, x3) = if₃(x1, x2, x3)
first₂(x1, x2) = first₂(x1, x2)
add₂(x1, x2) = add₂(x1, x2)
s₁(x1) = x1
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

          ↳ 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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

          ↳ 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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(first₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(first₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
add₂(x1, x2) = x1
if₃(x1, x2, x3) = x1
and₂(x1, x2) = 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(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = x1
add₂(x1, x2) = x1
and₂(x1, x2) = and₁(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(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
add₂(x1, x2) = x1
if₃(x1, x2, x3) = if₁(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(add₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(add₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
add₂(x1, x2) = add₁(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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(and₂(true, X)) -> mark₁(X)
active₁(and₂(false, Y)) -> mark₁(false)
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(first₂(0, X)) -> mark₁(nil)
active₁(first₂(s₁(X), cons₂(Y, Z))) -> mark₁(cons₂(Y, first₂(X, Z)))
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(and₂(X1, X2)) -> and₂(active₁(X1), X2)
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(active₁(X1), X2)
active₁(first₂(X1, X2)) -> first₂(X1, active₁(X2))
and₂(mark₁(X1), X2) -> mark₁(and₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
first₂(mark₁(X1), X2) -> mark₁(first₂(X1, X2))
first₂(X1, mark₁(X2)) -> mark₁(first₂(X1, X2))
proper₁(and₂(X1, X2)) -> and₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(first₂(X1, X2)) -> first₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
and₂(ok₁(X1), ok₁(X2)) -> ok₁(and₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
first₂(ok₁(X1), ok₁(X2)) -> ok₁(first₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(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.