Termination w.r.t. Q proof of /tmp/tmpPiEqUA/kabasci01.trs

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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

ACTIVE(length(cons(X, L))) → S(length(L))
ANY(proper(X)) → ANY(any(any(X)))
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
PROPER(eq(X1, X2)) → PROPER(X2)
ACTIVE(inf(X)) → INF(active(X))
PROPER(cons(any(X1), X2)) → ANY(any(proper(X1)))
PROPER(cons(any(X1), X2)) → CONS(any(any(proper(X1))), any(proper(X2)))
PROPER(length(X)) → LENGTH(proper(X))
ACTIVE(take(X1, X2)) → ACTIVE(X1)
PROPER(cons(any(X1), X2)) → ANY(proper(X2))
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(inf(X)) → INF(proper(X))
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(inf(X)) → S(X)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(inf(X)) → INF(s(X))
PROPER(cons(any(X1), X2)) → ANY(proper(X1))
TAKE(mark(X1), X2) → TAKE(X1, X2)
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
S(ok(X)) → S(X)
LENGTH(ok(X)) → LENGTH(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(inf(X)) → ACTIVE(X)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
ANY(proper(X)) → ANY(any(X))
TOP(mark(X)) → PROPER(X)
ACTIVE(length(cons(X, L))) → LENGTH(L)
ANY(X) → S(X)
ACTIVE(length(X)) → ACTIVE(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X2)
LENGTH(mark(X)) → LENGTH(X)
PROPER(inf(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(eq(X1, X2)) → EQ(proper(X1), proper(X2))
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
PROPER(eq(X1, X2)) → PROPER(X1)
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(length(X)) → LENGTH(active(X))
ANY(proper(X)) → ANY(X)
EQ(ok(X1), ok(X2)) → EQ(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
PROPER(take(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ACTIVE(length(cons(X, L))) → S(length(L))
ANY(proper(X)) → ANY(any(any(X)))
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
PROPER(eq(X1, X2)) → PROPER(X2)
ACTIVE(inf(X)) → INF(active(X))
PROPER(cons(any(X1), X2)) → ANY(any(proper(X1)))
PROPER(cons(any(X1), X2)) → CONS(any(any(proper(X1))), any(proper(X2)))
PROPER(length(X)) → LENGTH(proper(X))
ACTIVE(take(X1, X2)) → ACTIVE(X1)
PROPER(cons(any(X1), X2)) → ANY(proper(X2))
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(inf(X)) → INF(proper(X))
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(inf(X)) → S(X)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(inf(X)) → INF(s(X))
PROPER(cons(any(X1), X2)) → ANY(proper(X1))
TAKE(mark(X1), X2) → TAKE(X1, X2)
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
S(ok(X)) → S(X)
LENGTH(ok(X)) → LENGTH(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(inf(X)) → ACTIVE(X)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
ANY(proper(X)) → ANY(any(X))
TOP(mark(X)) → PROPER(X)
ACTIVE(length(cons(X, L))) → LENGTH(L)
ANY(X) → S(X)
ACTIVE(length(X)) → ACTIVE(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X2)
LENGTH(mark(X)) → LENGTH(X)
PROPER(inf(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(eq(X1, X2)) → EQ(proper(X1), proper(X2))
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
PROPER(eq(X1, X2)) → PROPER(X1)
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(length(X)) → LENGTH(active(X))
ANY(proper(X)) → ANY(X)
EQ(ok(X1), ok(X2)) → EQ(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
PROPER(take(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 10 SCCs with 26 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ANY(proper(X)) → ANY(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

ANY(proper(X)) → ANY(X)

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

ANY(proper(X)) → ANY(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LENGTH(ok(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LENGTH(ok(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(length(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(inf(X)) → PROPER(X)
PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(cons(any(X1), X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)
PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)

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

PROPER(cons(any(X1), X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(take(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(eq(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(length(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(take(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(inf(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(eq(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(cons(any(X1), X2)) → PROPER(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)

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

ACTIVE(take(X1, X2)) → ACTIVE(X2)
The graph contains the following edges 1 > 1
ACTIVE(inf(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(take(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(length(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(any(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(eq(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(inf(x₁)) = 2·x₁
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 2·x₁ + 2·x₂
POL(true) = 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ 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:

proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(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: Matrix interpretation [3]:
Non-tuple symbols:

M( eq(x₁, x₂) ) =

/	1	\
\	0	/

/	0	1	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( true ) =

/	0	\
\	0	/

M( mark(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( ok(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( any(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( take(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

/	1	0	\
\	0	0	/

x₂

M( 0 ) =

/	0	\
\	0	/

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( inf(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( false ) =

/	0	\
\	0	/

M( s(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	1	/

x₁

M( proper(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( length(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( nil ) =

/	0	\
\	0	/

Tuple symbols:

M( TOP(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

active(eq(X, Y)) → mark(false)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(take(0, X)) → mark(nil)
active(inf(X)) → mark(cons(X, inf(s(X))))
inf(ok(X)) → ok(inf(X))
inf(mark(X)) → mark(inf(X))
active(eq(0, 0)) → mark(true)
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
length(ok(X)) → ok(length(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(true) → ok(true)
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

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:

proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))

active(eq(0, 0)) → mark(true)
active(take(0, X)) → mark(nil)
active(length(nil)) → mark(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(eq(x₁, x₂)) = 2·x₁ + x₂
POL(false) = 0
POL(inf(x₁)) = 2·x₁
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 1
POL(ok(x₁)) = 2·x₁
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(true) = 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ 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(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
length(mark(X)) → mark(length(X))
length(ok(X)) → ok(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))

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(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
length(ok(X)) → ok(length(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))

Used ordering: POLO with Polynomial interpretation [25]:

POL(TOP(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(eq(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(false) = 1
POL(inf(x₁)) = x₁
POL(length(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(ok(x₁)) = 1 + 2·x₁
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 1 + 2·x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

active(inf(X)) → mark(cons(X, inf(s(X))))
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(length(X)) → length(active(X))
length(mark(X)) → mark(length(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
inf(mark(X)) → mark(inf(X))
inf(ok(X)) → ok(inf(X))
s(ok(X)) → ok(s(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.