Termination w.r.t. Q proof of /tmp/tmpDBAwUI/ExIntrod_GM01

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

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

active(head(cons(X, L))) → mark(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(head(x₁)) = 1 + x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(top(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

active(nats) → mark(adx(zeros))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 2
POL(nil) = 0
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(top(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

active(adx(nil)) → mark(nil)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 1
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = x₁
POL(top(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

active(tail(cons(X, L))) → mark(L)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(adx(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(head(x₁)) = x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nats) = 0
POL(nil) = 0
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(tail(x₁)) = 1 + 2·x₁
POL(top(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
INCR(ok(X)) → INCR(X)
PROPER(head(X)) → PROPER(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(incr(X)) → ACTIVE(X)
HEAD(mark(X)) → HEAD(X)
PROPER(incr(X)) → INCR(proper(X))
ACTIVE(adx(cons(X, L))) → ADX(L)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(adx(X)) → ADX(proper(X))
PROPER(tail(X)) → PROPER(X)
ACTIVE(zeros) → CONS(0, zeros)
TAIL(ok(X)) → TAIL(X)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → TAIL(active(X))
ADX(mark(X)) → ADX(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(incr(cons(X, L))) → S(X)
ADX(ok(X)) → ADX(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
PROPER(incr(X)) → PROPER(X)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
INCR(mark(X)) → INCR(X)
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(head(X)) → HEAD(proper(X))
PROPER(s(X)) → S(proper(X))
PROPER(tail(X)) → TAIL(proper(X))
TAIL(mark(X)) → TAIL(X)
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(tail(X)) → ACTIVE(X)
HEAD(ok(X)) → HEAD(X)
PROPER(adx(X)) → PROPER(X)
ACTIVE(adx(X)) → ADX(active(X))
ACTIVE(s(X)) → S(active(X))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
INCR(ok(X)) → INCR(X)
PROPER(head(X)) → PROPER(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(incr(X)) → ACTIVE(X)
HEAD(mark(X)) → HEAD(X)
PROPER(incr(X)) → INCR(proper(X))
ACTIVE(adx(cons(X, L))) → ADX(L)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(adx(X)) → ADX(proper(X))
PROPER(tail(X)) → PROPER(X)
ACTIVE(zeros) → CONS(0, zeros)
TAIL(ok(X)) → TAIL(X)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → TAIL(active(X))
ADX(mark(X)) → ADX(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(incr(cons(X, L))) → S(X)
ADX(ok(X)) → ADX(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
PROPER(incr(X)) → PROPER(X)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
INCR(mark(X)) → INCR(X)
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(head(X)) → HEAD(proper(X))
PROPER(s(X)) → S(proper(X))
PROPER(tail(X)) → TAIL(proper(X))
TAIL(mark(X)) → TAIL(X)
ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(tail(X)) → ACTIVE(X)
HEAD(ok(X)) → HEAD(X)
PROPER(adx(X)) → PROPER(X)
ACTIVE(adx(X)) → ADX(active(X))
ACTIVE(s(X)) → S(active(X))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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 [15,17,22] contains 9 SCCs with 21 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

TAIL(ok(X)) → TAIL(X)
The graph contains the following edges 1 > 1
TAIL(mark(X)) → TAIL(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

HEAD(mark(X)) → HEAD(X)
HEAD(ok(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

HEAD(mark(X)) → HEAD(X)
HEAD(ok(X)) → HEAD(X)

From the DPs we obtained the following set of size-change graphs:

HEAD(mark(X)) → HEAD(X)
The graph contains the following edges 1 > 1
HEAD(ok(X)) → HEAD(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

ADX(mark(X)) → ADX(X)
ADX(ok(X)) → ADX(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

ADX(mark(X)) → ADX(X)
ADX(ok(X)) → ADX(X)

From the DPs we obtained the following set of size-change graphs:

ADX(mark(X)) → ADX(X)
The graph contains the following edges 1 > 1
ADX(ok(X)) → ADX(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ 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)

From the DPs we obtained the following set of size-change graphs:

S(ok(X)) → S(X)
The graph contains the following edges 1 > 1
S(mark(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ 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)

From the DPs we obtained the following set of size-change graphs:

CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

INCR(mark(X)) → INCR(X)
INCR(ok(X)) → INCR(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

INCR(mark(X)) → INCR(X)
INCR(ok(X)) → INCR(X)

From the DPs we obtained the following set of size-change graphs:

INCR(mark(X)) → INCR(X)
The graph contains the following edges 1 > 1
INCR(ok(X)) → INCR(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

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

PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(tail(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(adx(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

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

PROPER(incr(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(tail(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(adx(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)

From the DPs we obtained the following set of size-change graphs:

PROPER(incr(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(tail(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(adx(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(head(X)) → PROPER(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → ACTIVE(X)

From the DPs we obtained the following set of size-change graphs:

ACTIVE(incr(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(s(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(adx(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(head(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(tail(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
adx(mark(X)) → mark(adx(X))
adx(ok(X)) → ok(adx(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
incr(ok(X)) → ok(incr(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.

TOP(ok(X)) → TOP(active(X))

Used ordering: Combined order from the following AFS and order.
TOP(x1) = TOP(x1)
mark(x1) = mark(x1)
proper(x1) = x1
ok(x1) = x1
active(x1) = x1
tail(x1) = x1
head(x1) = head(x1)
s(x1) = s(x1)
adx(x1) = adx(x1)
cons(x1, x2) = cons(x1)
incr(x1) = incr(x1)
zeros = zeros
0 = 0
nil = nil
nats = nats

Recursive path order with status [2].
Quasi-Precedence:

head₁ > mark₁
adx₁ > [incr₁, nil] > s₁ > mark₁
adx₁ > [incr₁, nil] > cons₁ > mark₁
zeros > cons₁ > mark₁
zeros > 0

Status:

cons₁: [1]
adx₁: [1]
incr₁: [1]
zeros: multiset
head₁: multiset
mark₁: [1]
TOP₁: [1]
s₁: [1]
nats: multiset
0: multiset
nil: multiset

The following usable rules [17] were oriented:

tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
proper(0) → ok(0)
proper(zeros) → ok(zeros)
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(incr(X)) → incr(proper(X))
incr(ok(X)) → ok(incr(X))
incr(mark(X)) → mark(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(ok(X)) → ok(s(X))
s(mark(X)) → mark(s(X))
adx(ok(X)) → ok(adx(X))
adx(mark(X)) → mark(adx(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
adx(mark(X)) → mark(adx(X))
adx(ok(X)) → ok(adx(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
incr(ok(X)) → ok(incr(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ RuleRemovalProof

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
tail(mark(X)) → mark(tail(X))
tail(ok(X)) → ok(tail(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
adx(mark(X)) → mark(adx(X))
adx(ok(X)) → ok(adx(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
incr(ok(X)) → ok(incr(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

TOP(ok(X)) → TOP(active(X))

Strictly oriented rules of the TRS R:

active(tail(X)) → tail(active(X))
tail(ok(X)) → ok(tail(X))
head(ok(X)) → ok(head(X))
adx(ok(X)) → ok(adx(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(adx(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(head(x₁)) = 2·x₁
POL(incr(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(ok(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 2·x₁
POL(tail(x₁)) = 1 + 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ RuleRemovalProof

                                          ↳ QDP

                                            ↳ PisEmptyProof

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(zeros) → mark(cons(0, zeros))
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
tail(mark(X)) → mark(tail(X))
head(mark(X)) → mark(head(X))
adx(mark(X)) → mark(adx(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
incr(ok(X)) → ok(incr(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.