Termination w.r.t. Q proof of TRS/TRCSREx15_Luc98

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 10 SCCs with 21 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

FROM₁(ok₁(X)) -> FROM₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ok₁(x₁) ) = 3x₁ + 3

POL( FROM₁(x₁) ) = 2x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = 3x₂ + 3

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

S₁(ok₁(X)) -> S₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( S₁(x₁) ) = 2x₁ + 3

POL( ok₁(x₁) ) = 3x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST₂(x₁, x₂) ) = 3x₂ + 2

POL( mark₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST₂(x₁, x₂) ) = 2x₁ + 3x₂ + 2

POL( mark₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ADD₂(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( mark₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IF₃(x₁, ..., x₃) ) = 2x₁ + 2x₃ + 1

POL( mark₁(x₁) ) = x₁

POL( ok₁(x₁) ) = 2x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = 2x₁ + 1

POL( IF₃(x₁, ..., x₃) ) = 2x₁ + 2x₂ + 3x₃ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( AND₂(x₁, x₂) ) = x₁ + 2x₂ + 2

POL( mark₁(x₁) ) = 3x₁ + 3

POL( ok₁(x₁) ) = 2x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = 2x₁

POL( if₃(x₁, ..., x₃) ) = x₁ + 2x₂ + 2x₃ + 1

POL( and₂(x₁, x₂) ) = x₁ + 2x₂ + 3

POL( first₂(x₁, x₂) ) = 2x₁ + 2x₂ + 1

POL( PROPER₁(x₁) ) = x₁ + 3

POL( add₂(x₁, x₂) ) = 2x₁ + 2x₂ + 1

POL( s₁(x₁) ) = 2x₁

POL( cons₂(x₁, x₂) ) = x₁ + 2x₂ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = 2x₁ + 2

POL( s₁(x₁) ) = 3x₁ + 1

POL( PROPER₁(x₁) ) = 2x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( if₃(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 3

POL( and₂(x₁, x₂) ) = x₁

POL( ACTIVE₁(x₁) ) = x₁ + 2

POL( first₂(x₁, x₂) ) = 2x₁ + 2x₂ + 1

POL( add₂(x₁, x₂) ) = x₁ + 3x₂ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE₁(and₂(X1, X2)) -> ACTIVE₁(X1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVE₁(x₁) ) = 2x₁ + 2

POL( and₂(x₁, x₂) ) = 3x₁ + x₂ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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

            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The remaining pairs can at least be oriented weakly.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = 2x₁ + 2

POL( mark₁(x₁) ) = x₁ + 1

POL( if₃(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 2

POL( first₂(x₁, x₂) ) = 2x₁ + 2x₂ + 1

POL( active₁(x₁) ) = x₁

POL( 0 ) = 2

POL( nil ) = 3

POL( cons₂(x₁, x₂) ) = max{0, -3}

POL( true ) = 1

POL( and₂(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( false ) = 0

POL( add₂(x₁, x₂) ) = 3x₁ + 2x₂ + 1

POL( s₁(x₁) ) = x₁ + 1

POL( TOP₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = x₁

POL( proper₁(x₁) ) = x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

TOP₁(ok₁(X)) -> TOP₁(active₁(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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from₁(x₁) ) = 3x₁ + 2

POL( mark₁(x₁) ) = max{0, -3}

POL( if₃(x₁, ..., x₃) ) = 3x₂ + 3x₃ + 3

POL( first₂(x₁, x₂) ) = 3x₁ + 3

POL( active₁(x₁) ) = x₁

POL( 0 ) = 1

POL( nil ) = 2

POL( cons₂(x₁, x₂) ) = 2x₂

POL( true ) = 0

POL( and₂(x₁, x₂) ) = x₁

POL( false ) = 0

POL( add₂(x₁, x₂) ) = 3x₁ + 3x₂ + 1

POL( s₁(x₁) ) = max{0, -3}

POL( TOP₁(x₁) ) = max{0, 3x₁ - 3}

POL( ok₁(x₁) ) = 3x₁ + 3

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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.