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:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(g(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(f(X)) → G(X)
ACTIVE(g(s(X))) → G(X)
ACTIVE(g(s(X))) → S(g(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
S(mark(X)) → S(X)
G(mark(X)) → G(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
PROPER(f(X)) → PROPER(X)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
SEL(X1, mark(X2)) → SEL(X1, X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
S(ok(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
ACTIVE(f(X)) → F(active(X))
ACTIVE(g(s(X))) → S(s(g(X)))
PROPER(sel(X1, X2)) → PROPER(X2)
TOP(ok(X)) → ACTIVE(X)
SEL(mark(X1), X2) → SEL(X1, X2)
F(mark(X)) → F(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
ACTIVE(g(X)) → G(active(X))
ACTIVE(f(X)) → F(g(X))
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(g(0)) → S(0)
G(ok(X)) → G(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(f(X)) → CONS(X, f(g(X)))
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(g(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(f(X)) → G(X)
ACTIVE(g(s(X))) → G(X)
ACTIVE(g(s(X))) → S(g(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
S(mark(X)) → S(X)
G(mark(X)) → G(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
PROPER(f(X)) → PROPER(X)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
SEL(X1, mark(X2)) → SEL(X1, X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
S(ok(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
ACTIVE(f(X)) → F(active(X))
ACTIVE(g(s(X))) → S(s(g(X)))
PROPER(sel(X1, X2)) → PROPER(X2)
TOP(ok(X)) → ACTIVE(X)
SEL(mark(X1), X2) → SEL(X1, X2)
F(mark(X)) → F(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
ACTIVE(g(X)) → G(active(X))
ACTIVE(f(X)) → F(g(X))
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(g(0)) → S(0)
G(ok(X)) → G(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(f(X)) → CONS(X, f(g(X)))
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(g(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(f(X)) → G(X)
ACTIVE(g(s(X))) → G(X)
ACTIVE(g(s(X))) → S(g(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
S(mark(X)) → S(X)
G(mark(X)) → G(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
PROPER(f(X)) → PROPER(X)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(s(X)) → S(proper(X))
TOP(mark(X)) → TOP(proper(X))
ACTIVE(s(X)) → ACTIVE(X)
S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
PROPER(g(X)) → G(proper(X))
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
PROPER(g(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
ACTIVE(f(X)) → F(active(X))
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
ACTIVE(g(s(X))) → S(s(g(X)))
PROPER(sel(X1, X2)) → PROPER(X2)
TOP(ok(X)) → ACTIVE(X)
SEL(mark(X1), X2) → SEL(X1, X2)
F(mark(X)) → F(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
ACTIVE(f(X)) → F(g(X))
ACTIVE(g(X)) → G(active(X))
PROPER(s(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
PROPER(cons(X1, X2)) → PROPER(X2)
F(ok(X)) → F(X)
ACTIVE(g(0)) → S(0)
G(ok(X)) → G(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(f(X)) → CONS(X, f(g(X)))
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


SEL(ok(X1), ok(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x2
mark(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


SEL(X1, mark(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.

SEL(mark(X1), X2) → SEL(X1, X2)
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


SEL(mark(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ 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:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.

S(ok(X)) → S(X)
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


S(ok(X)) → S(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

G(mark(X)) → G(X)
G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


G(ok(X)) → G(X)
The remaining pairs can at least be oriented weakly.

G(mark(X)) → G(X)
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
mark(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


G(mark(X)) → G(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.

CONS(mark(X1), X2) → CONS(X1, X2)
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x2
ok(x1)  =  ok(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

F(mark(X)) → F(X)
F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


F(ok(X)) → F(X)
The remaining pairs can at least be oriented weakly.

F(mark(X)) → F(X)
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
mark(x1)  =  x1
ok(x1)  =  ok(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


F(mark(X)) → F(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.

PROPER(g(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
g(x1)  =  x1
s(x1)  =  x1
f(x1)  =  x1
sel(x1, x2)  =  sel(x1, x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(s(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
g(x1)  =  x1
f(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(f(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.

PROPER(g(X)) → PROPER(X)
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
g(x1)  =  x1
f(x1)  =  f(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


PROPER(g(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
g(x1)  =  g(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
sel(x1, x2)  =  sel(x1, x2)
g(x1)  =  x1
f(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(s(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
g(x1)  =  x1
f(x1)  =  x1
cons(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
g(x1)  =  x1
f(x1)  =  x1
cons(x1, x2)  =  cons(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(f(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.

ACTIVE(g(X)) → ACTIVE(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
g(x1)  =  x1
f(x1)  =  f(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
              ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


ACTIVE(g(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
g(x1)  =  g(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ QDP
                                    ↳ PisEmptyProof
              ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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