NO Termination w.r.t. Q proof of ../tpdb/TRS/TRCSR/OvConsOS_nosorts_GM.trs
Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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

a__and(tt, X) → mark(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__length(x1)) = x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(tt) = 1   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.

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

a__zeros → cons(0, zeros)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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

a__take(0, IL) → nil
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__and(x1, x2)) = 2·x1 + 2·x2   
POL(a__length(x1)) = x1   
POL(a__take(x1, x2)) = 1 + 2·x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 2·x1 + 2·x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + 2·x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.

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

a__zeros → cons(0, zeros)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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

a__length(nil) → 0
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__and(x1, x2)) = 2·x1 + x2   
POL(a__length(x1)) = x1   
POL(a__take(x1, x2)) = x1 + 2·x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 2   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(tt) = 2   
POL(zeros) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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:

MARK(take(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(zeros) → A__ZEROS
MARK(take(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(length(X)) → MARK(X)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
A__TAKE(s(M), cons(N, IL)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(zeros) → A__ZEROS
MARK(take(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(length(X)) → MARK(X)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
A__TAKE(s(M), cons(N, IL)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
A__TAKE(s(M), cons(N, IL)) → MARK(N)
MARK(take(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(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:

MARK(and(X1, X2)) → MARK(X1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__LENGTH(x1)) = x1   
POL(A__TAKE(x1, x2)) = x1 + x2   
POL(MARK(x1)) = x1   
POL(a__and(x1, x2)) = 1 + x1 + x2   
POL(a__length(x1)) = 2·x1   
POL(a__take(x1, x2)) = 2·x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
A__TAKE(s(M), cons(N, IL)) → MARK(N)
MARK(take(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(length(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(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:

MARK(take(X1, X2)) → MARK(X2)
A__TAKE(s(M), cons(N, IL)) → MARK(N)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__LENGTH(x1)) = 1 + 2·x1   
POL(A__TAKE(x1, x2)) = 2 + x1 + 2·x2   
POL(MARK(x1)) = 1 + 2·x1   
POL(a__and(x1, x2)) = x1 + x2   
POL(a__length(x1)) = 2·x1   
POL(a__take(x1, x2)) = 1 + 2·x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + 2·x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof

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

MARK(s(X)) → MARK(X)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(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:

MARK(length(X)) → A__LENGTH(mark(X))
MARK(length(X)) → MARK(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__LENGTH(x1)) = x1   
POL(MARK(x1)) = x1   
POL(a__and(x1, x2)) = 2·x1 + x2   
POL(a__length(x1)) = 2 + x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(length(x1)) = 2 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof

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

MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → MARK(L)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                      ↳ QDP

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QDPSizeChangeProof
                                      ↳ QDP

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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: 



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__LENGTH(cons(N, L)) → A__LENGTH(mark(L)) at position [0] we obtained the following new rules:

A__LENGTH(cons(y0, nil)) → A__LENGTH(nil)
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, 0)) → A__LENGTH(0)
A__LENGTH(cons(y0, tt)) → A__LENGTH(tt)
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, s(x0))) → A__LENGTH(s(mark(x0)))
A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof

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

A__LENGTH(cons(y0, nil)) → A__LENGTH(nil)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, 0)) → A__LENGTH(0)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, tt)) → A__LENGTH(tt)
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, s(x0))) → A__LENGTH(s(mark(x0)))
A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ RuleRemovalProof

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

A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(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:

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__LENGTH(x1)) = x1   
POL(a__and(x1, x2)) = 1 + 2·x1 + x2   
POL(a__length(x1)) = x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zeros) = 0   
POL(and(x1, x2)) = 1 + 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(tt) = 2   
POL(zeros) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ RuleRemovalProof
                                                  ↳ QDP
                                                    ↳ QDPOrderProof

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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.


A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
The remaining pairs can at least be oriented weakly.

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(A__LENGTH(x1)) = x1   
POL(a__and(x1, x2)) = x2   
POL(a__length(x1)) = 0   
POL(a__take(x1, x2)) = 1   
POL(a__zeros) = 1   
POL(and(x1, x2)) = x2   
POL(cons(x1, x2)) = 1   
POL(length(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = 0   
POL(take(x1, x2)) = 1   
POL(tt) = 0   
POL(zeros) = 0   

The following usable rules [17] were oriented:

a__zeros → zeros
mark(length(X)) → a__length(mark(X))
a__zeros → cons(0, zeros)
a__and(X1, X2) → and(X1, X2)
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ RuleRemovalProof
                                                  ↳ QDP
                                                    ↳ QDPOrderProof
                                                      ↳ QDP
                                                        ↳ QDPOrderProof

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

A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

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.


A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
The remaining pairs can at least be oriented weakly.

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( mark(x1) ) =
/0\
\1/
+
/01\
\01/
·x1

M( zeros ) =
/0\
\0/

M( length(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( a__length(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( a__zeros ) =
/0\
\1/

M( 0 ) =
/0\
\0/

M( nil ) =
/0\
\0/

M( cons(x1, x2) ) =
/0\
\1/
+
/00\
\00/
·x1+
/01\
\01/
·x2

M( a__take(x1, x2) ) =
/0\
\1/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( tt ) =
/0\
\0/

M( a__and(x1, x2) ) =
/1\
\1/
+
/00\
\00/
·x1+
/10\
\10/
·x2

M( and(x1, x2) ) =
/0\
\1/
+
/00\
\00/
·x1+
/10\
\10/
·x2

M( s(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( take(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

Tuple symbols:
M( A__LENGTH(x1) ) = 0+
[1,0]
·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:

a__zeros → zeros
mark(length(X)) → a__length(mark(X))
a__zeros → cons(0, zeros)
a__and(X1, X2) → and(X1, X2)
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ AND
                                      ↳ QDP
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ QDP
                                                ↳ RuleRemovalProof
                                                  ↳ QDP
                                                    ↳ QDPOrderProof
                                                      ↳ QDP
                                                        ↳ QDPOrderProof
                                                          ↳ QDP
                                                            ↳ NonTerminationProof

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)
a__take(X1, X2) → take(X1, X2)


s = A__LENGTH(a__zeros) evaluates to t =A__LENGTH(a__zeros)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

A__LENGTH(a__zeros) → A__LENGTH(cons(0, zeros))
with rule a__zeros → cons(0, zeros) at position [0] and matcher [ ]

A__LENGTH(cons(0, zeros)) → A__LENGTH(a__zeros)
with rule A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.