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:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

tail(cons(X, XS)) → activate(XS)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zip(x1, x2)) = x1 + 2·x2   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 1 + 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + 2·x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

zip(nil, XS) → nil
zip(X, nil) → nil
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zip(x1, x2)) = 1 + 2·x1 + x2   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 1 + 2·x1 + x2   




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

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

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

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = 2·x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 1 + 2·x1   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zip(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + x2   




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

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

take(0, XS) → nil
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(n__cons(x1, x2)) = 2·x1 + x2   
POL(n__incr(x1)) = x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(n__zip(x1, x2)) = 2·x1 + 2·x2   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   




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

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

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

TAKE(s(N), cons(X, XS)) → CONS(X, n__take(N, activate(XS)))
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
INCR(cons(X, XS)) → CONS(s(X), n__incr(activate(XS)))
PAIRNS → CONS(0, n__incr(n__oddNs))
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__repItems(X)) → ACTIVATE(X)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ODDNS → INCR(pairNs)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ODDNS → PAIRNS
ZIP(cons(X, XS), cons(Y, YS)) → CONS(pair(X, Y), n__zip(activate(XS), activate(YS)))
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__oddNs) → ODDNS

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

TAKE(s(N), cons(X, XS)) → CONS(X, n__take(N, activate(XS)))
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
INCR(cons(X, XS)) → CONS(s(X), n__incr(activate(XS)))
PAIRNS → CONS(0, n__incr(n__oddNs))
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__repItems(X)) → ACTIVATE(X)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ODDNS → INCR(pairNs)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ODDNS → PAIRNS
ZIP(cons(X, XS), cons(Y, YS)) → CONS(pair(X, Y), n__zip(activate(XS), activate(YS)))
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__oddNs) → ODDNS

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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 7 less nodes.

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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ODDNS → INCR(pairNs)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__repItems(X)) → ACTIVATE(X)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__oddNs) → ODDNS
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__repItems(X)) → REPITEMS(activate(X))
ACTIVATE(n__repItems(X)) → ACTIVATE(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(INCR(x1)) = x1   
POL(ODDNS) = 0   
POL(REPITEMS(x1)) = 2·x1   
POL(TAKE(x1, x2)) = x1 + x2   
POL(ZIP(x1, x2)) = 2·x1 + x2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(n__cons(x1, x2)) = 2·x1 + x2   
POL(n__incr(x1)) = x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2 + 2·x1   
POL(n__take(x1, x2)) = 2·x1 + x2   
POL(n__zip(x1, x2)) = 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(zip(x1, x2)) = 2·x1 + x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ODDNS → INCR(pairNs)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__oddNs) → ODDNS
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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 1 less node.

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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ODDNS → INCR(pairNs)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__oddNs) → ODDNS
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X1)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__zip(X1, X2)) → ZIP(activate(X1), activate(X2))
ACTIVATE(n__zip(X1, X2)) → ACTIVATE(X2)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(INCR(x1)) = 2·x1   
POL(ODDNS) = 0   
POL(TAKE(x1, x2)) = 2·x1 + 2·x2   
POL(ZIP(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = 2·x1 + x2   
POL(n__zip(x1, x2)) = 1 + 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(zip(x1, x2)) = 1 + 2·x1 + x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ODDNS → INCR(pairNs)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(INCR(x1)) = x1   
POL(ODDNS) = 0   
POL(TAKE(x1, x2)) = x1 + x2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(n__cons(x1, x2)) = 2·x1 + x2   
POL(n__incr(x1)) = x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = 1 + x1 + 2·x2   
POL(n__zip(x1, x2)) = 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + 2·x2   
POL(zip(x1, x2)) = 2·x1 + x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDNS → INCR(pairNs)
ACTIVATE(n__oddNs) → ODDNS
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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 1 less node.

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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ODDNS → INCR(pairNs)
ACTIVATE(n__oddNs) → ODDNS

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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

ODDNS → INCR(cons(0, n__incr(n__oddNs)))



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ACTIVATE(n__incr(X)) → INCR(activate(X)) at position [0] we obtained the following new rules:

ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ODDNS → INCR(cons(0, n__incr(n__oddNs)))
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__incr(n__take(x0, x1))) → INCR(take(activate(x0), activate(x1)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(INCR(x1)) = 2·x1   
POL(ODDNS) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = 1 + x1 + x2   
POL(n__zip(x1, x2)) = 2·x1 + 2·x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + 2·x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__oddNs) → ODDNS
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__incr(n__zip(x0, x1))) → INCR(zip(activate(x0), activate(x1)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(INCR(x1)) = 2·x1   
POL(ODDNS) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 2·x1   
POL(n__take(x1, x2)) = x1 + 2·x2   
POL(n__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → 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:

ACTIVATE(n__incr(n__repItems(x0))) → INCR(repItems(activate(x0)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(INCR(x1)) = 2·x1   
POL(ODDNS) = 0   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__incr(x1)) = 2·x1   
POL(n__oddNs) = 0   
POL(n__repItems(x1)) = 1 + 2·x1   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zip(x1, x2)) = 2·x1 + x2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 2·x1 + x2   



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

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

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:

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__oddNs)) → INCR(oddNs)
ACTIVATE(n__incr(n__cons(x0, x1))) → INCR(cons(activate(x0), x1))
ACTIVATE(n__oddNs) → ODDNS
ODDNS → INCR(cons(0, n__incr(n__oddNs)))

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(n__oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNs → n__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X


s = ACTIVATE(n__incr(n__oddNs)) evaluates to t =ACTIVATE(n__incr(n__oddNs))

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



Rewriting sequence

ACTIVATE(n__incr(n__oddNs)) → ACTIVATE(n__oddNs)
with rule ACTIVATE(n__incr(X)) → ACTIVATE(X) at position [] and matcher [X / n__oddNs]

ACTIVATE(n__oddNs) → ODDNS
with rule ACTIVATE(n__oddNs) → ODDNS at position [] and matcher [ ]

ODDNS → INCR(cons(0, n__incr(n__oddNs)))
with rule ODDNS → INCR(cons(0, n__incr(n__oddNs))) at position [] and matcher [ ]

INCR(cons(0, n__incr(n__oddNs))) → ACTIVATE(n__incr(n__oddNs))
with rule INCR(cons(X, XS)) → ACTIVATE(XS)

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.