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:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.


QTRS
  ↳ RRRPoloQTRSProof

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

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

POL(0) = 0   
POL(a__incr(x1)) = x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = x1   
POL(a__take(x1, x2)) = 2·x1 + x2   
POL(a__zip(x1, x2)) = 2 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
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(tail(x1)) = x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(zip(x1, x2)) = 2 + x1 + x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
QTRS
      ↳ RRRPoloQTRSProof

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

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

POL(0) = 0   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = x1   
POL(a__take(x1, x2)) = 2 + x1 + x2   
POL(a__zip(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
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(tail(x1)) = x1   
POL(take(x1, x2)) = 2 + x1 + x2   
POL(zip(x1, x2)) = x1 + x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
QTRS
          ↳ RRRPoloQTRSProof

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

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

POL(0) = 0   
POL(a__incr(x1)) = x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 2·x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zip(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 2   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(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:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

a__tail(cons(X, XS)) → mark(XS)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 1 + x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zip(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
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 + x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + x2   




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

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

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(take(X1, X2)) → MARK(X2)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(repItems(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → A__PAIRNS
MARK(pair(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)
MARK(oddNs) → A__ODDNS
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
A__ODDNSA__PAIRNS
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(X1, X2)) → MARK(X1)
A__ODDNSA__INCR(a__pairNs)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(take(X1, X2)) → MARK(X2)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(repItems(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → A__PAIRNS
MARK(pair(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)
MARK(oddNs) → A__ODDNS
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
A__ODDNSA__PAIRNS
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(X1, X2)) → MARK(X1)
A__ODDNSA__INCR(a__pairNs)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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 3 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:

MARK(take(X1, X2)) → MARK(X2)
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(repItems(X)) → MARK(X)
MARK(s(X)) → MARK(X)
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → MARK(X1)
A__ODDNSA__INCR(a__pairNs)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(take(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(repItems(X)) → MARK(X)
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = 2·x1   
POL(A__ODDNS) = 0   
POL(A__REPITEMS(x1)) = 1 + 2·x1   
POL(A__TAKE(x1, x2)) = x1 + x2   
POL(A__ZIP(x1, x2)) = x1 + 2·x2   
POL(MARK(x1)) = 2·x1   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 1 + 2·x1   
POL(a__tail(x1)) = x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zip(x1, x2)) = x1 + 2·x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + 2·x2   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ 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(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → MARK(X1)
A__ODDNSA__INCR(a__pairNs)
MARK(take(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(tail(X)) → MARK(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = 2·x1   
POL(A__ODDNS) = 0   
POL(A__TAKE(x1, x2)) = x1 + x2   
POL(A__ZIP(x1, x2)) = x1 + x2   
POL(MARK(x1)) = 2·x1   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 1 + x1   
POL(a__take(x1, x2)) = x1 + 2·x2   
POL(a__zip(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
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 + x1   
POL(take(x1, x2)) = 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
                              ↳ RuleRemovalProof
QDP
                                  ↳ RuleRemovalProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → MARK(X1)
A__ODDNSA__INCR(a__pairNs)
MARK(take(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X2)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = 2 + x1   
POL(A__ODDNS) = 2   
POL(A__TAKE(x1, x2)) = 2 + x1 + x2   
POL(A__ZIP(x1, x2)) = 2 + x1 + x2   
POL(MARK(x1)) = 2 + x1   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 2·x1   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zip(x1, x2)) = 2 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
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(tail(x1)) = 2·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
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
QDP
                                      ↳ DependencyGraphProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
A__ODDNSA__INCR(a__pairNs)
MARK(take(X1, X2)) → MARK(X1)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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 2 less nodes.

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

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

MARK(take(X1, X2)) → MARK(X2)
MARK(incr(X)) → A__INCR(mark(X))
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
A__ODDNSA__INCR(a__pairNs)
MARK(take(X1, X2)) → MARK(X1)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = 2·x1   
POL(A__ODDNS) = 0   
POL(A__TAKE(x1, x2)) = x1 + x2   
POL(MARK(x1)) = x1   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 2·x1   
POL(a__take(x1, x2)) = 1 + x1 + x2   
POL(a__zip(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
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)) = 2·x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 2·x1 + x2   



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

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(pair(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
A__ODDNSA__INCR(a__pairNs)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
QDP
                                                  ↳ RuleRemovalProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(pair(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
A__ODDNSA__INCR(a__pairNs)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
MARK(oddNs) → A__ODDNS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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:

MARK(oddNs) → A__ODDNS


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = x1   
POL(A__ODDNS) = 2   
POL(MARK(x1)) = 2·x1   
POL(a__incr(x1)) = x1   
POL(a__oddNs) = 2   
POL(a__pairNs) = 2   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = x1   
POL(a__take(x1, x2)) = x1 + 2·x2   
POL(a__zip(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 2   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 2   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(take(x1, x2)) = 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
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
QDP
                                                      ↳ DependencyGraphProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
A__ODDNSA__INCR(a__pairNs)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
QDP
                                                          ↳ QDPOrderProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(pair(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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.


MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(pair(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__INCR(cons(X, XS)) → MARK(X)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = x1   
POL(MARK(x1)) = x1   
POL(a__incr(x1)) = 1 + x1   
POL(a__oddNs) = 1   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = x1   
POL(a__tail(x1)) = 0   
POL(a__take(x1, x2)) = x1 + x2   
POL(a__zip(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1   
POL(incr(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 1   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 0   
POL(take(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 1 + x1 + x2   

The following usable rules [17] were oriented:

mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__pairNscons(0, incr(oddNs))
a__repItems(X) → repItems(X)
a__tail(X) → tail(X)
a__zip(X1, X2) → zip(X1, X2)
a__take(X1, X2) → take(X1, X2)
a__oddNsoddNs
a__incr(X) → incr(X)
a__pairNspairNs
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)



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

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

MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X1)
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                                        ↳ QDP
                                                          ↳ QDPOrderProof
                                                            ↳ QDP
                                                              ↳ DependencyGraphProof
QDP
                                                                  ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

a__pairNscons(0, incr(oddNs))
a__oddNsa__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNspairNs
a__incr(X) → incr(X)
a__oddNsoddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                                        ↳ QDP
                                                          ↳ QDPOrderProof
                                                            ↳ QDP
                                                              ↳ DependencyGraphProof
                                                                ↳ QDP
                                                                  ↳ UsableRulesProof
QDP
                                                                      ↳ QDPSizeChangeProof

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

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