Termination w.r.t. Q proof of TRS/TRCSREx4_Zan97

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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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:

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(f₁(X)) -> ACTIVE₁(X)
ACTIVE₁(f₁(X)) -> G₁(X)
ACTIVE₁(g₁(s₁(X))) -> G₁(X)
ACTIVE₁(g₁(s₁(X))) -> S₁(g₁(X))
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
S₁(mark₁(X)) -> S₁(X)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
PROPER₁(f₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(f₁(X)) -> F₁(active₁(X))
ACTIVE₁(g₁(s₁(X))) -> S₁(s₁(g₁(X)))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
F₁(mark₁(X)) -> F₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
ACTIVE₁(f₁(X)) -> F₁(g₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
F₁(ok₁(X)) -> F₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
ACTIVE₁(g₁(0)) -> S₁(0)
G₁(ok₁(X)) -> G₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(f₁(X)) -> CONS₂(X, f₁(g₁(X)))
PROPER₁(f₁(X)) -> F₁(proper₁(X))

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(f₁(X)) -> ACTIVE₁(X)
ACTIVE₁(f₁(X)) -> G₁(X)
ACTIVE₁(g₁(s₁(X))) -> G₁(X)
ACTIVE₁(g₁(s₁(X))) -> S₁(g₁(X))
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
S₁(mark₁(X)) -> S₁(X)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
PROPER₁(f₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(f₁(X)) -> F₁(active₁(X))
ACTIVE₁(g₁(s₁(X))) -> S₁(s₁(g₁(X)))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
F₁(mark₁(X)) -> F₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
ACTIVE₁(f₁(X)) -> F₁(g₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
F₁(ok₁(X)) -> F₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
ACTIVE₁(g₁(0)) -> S₁(0)
G₁(ok₁(X)) -> G₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(f₁(X)) -> CONS₂(X, f₁(g₁(X)))
PROPER₁(f₁(X)) -> F₁(proper₁(X))

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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 8 SCCs with 21 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SEL₂(x₁, x₂)) = 3·x₁ + 3·x₁·x₂ + 3·x₂
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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)
S₁(mark₁(X)) -> S₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S₁(x₁)) = 3·x₁ + 3·x₁²
POL(mark₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(ok₁(x₁)) = 3 + x₁ + 2·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

G₁(mark₁(X)) -> G₁(X)
G₁(ok₁(X)) -> G₁(X)

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

G₁(mark₁(X)) -> G₁(X)
G₁(ok₁(X)) -> G₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G₁(x₁)) = 3·x₁ + 3·x₁²
POL(mark₁(x₁)) = 3 + x₁ + 2·x₁²
POL(ok₁(x₁)) = 3 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

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

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(mark₁(X1), X2) -> CONS₂(X1, X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS₂(x₁, x₂)) = 3·x₁ + 3·x₁·x₂ + 3·x₂
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

F₁(mark₁(X)) -> F₁(X)
F₁(ok₁(X)) -> F₁(X)

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

F₁(mark₁(X)) -> F₁(X)
F₁(ok₁(X)) -> F₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₁(x₁)) = 3·x₁ + 3·x₁²
POL(mark₁(x₁)) = 3 + x₁ + 2·x₁²
POL(ok₁(x₁)) = 3 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(f₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(g₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(f₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = 3·x₁ + 3·x₁²
POL(cons₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(f₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(g₁(x₁)) = 3 + x₁
POL(s₁(x₁)) = 3 + 3·x₁ + 2·x₁²
POL(sel₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
ACTIVE₁(f₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
ACTIVE₁(f₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = 3·x₁ + 3·x₁²
POL(cons₂(x₁, x₂)) = 3 + 2·x₁
POL(f₁(x₁)) = 3 + 2·x₁
POL(g₁(x₁)) = 3 + x₁
POL(s₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(sel₂(x₁, x₂)) = 3 + 2·x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active₁(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

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(f₁(X)) -> mark₁(cons₂(X, f₁(g₁(X))))
active₁(g₁(0)) -> mark₁(s₁(0))
active₁(g₁(s₁(X))) -> mark₁(s₁(s₁(g₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(f₁(X)) -> f₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
f₁(mark₁(X)) -> mark₁(f₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
f₁(ok₁(X)) -> ok₁(f₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
g₁(ok₁(X)) -> ok₁(g₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.