YES Termination proof of ../tpdb/TRS/CSR/Ex5_DLMMU04.trs
Termination of the following Term Rewriting System could be proven:

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

pairNs → cons(0, incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incr(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS)) → XS
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, cons(X, repItems(XS)))

The replacement map contains the following entries:

pairNs: empty set
cons: {1}
0: empty set
incr: {1}
oddNs: empty set
s: {1}
take: {1, 2}
nil: empty set
zip: {1, 2}
pair: {1, 2}
tail: {1}
repItems: {1}


↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incr(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS)) → XS
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, cons(X, repItems(XS)))

The replacement map contains the following entries:

pairNs: empty set
cons: {1}
0: empty set
incr: {1}
oddNs: empty set
s: {1}
take: {1, 2}
nil: empty set
zip: {1, 2}
pair: {1, 2}
tail: {1}
repItems: {1}

We applied the Zantema [34] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
tail(cons(X, XS)) → a(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
take(x1, x2) → takeInact(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
tail(cons(X, XS)) → a(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
take(x1, x2) → takeInact(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

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

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

POL(0) = 0   
POL(a(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(consInact(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = 2·x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(takeInact(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipInact(x1, x2)) = 2·x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
take(x1, x2) → takeInact(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
take(x1, x2) → takeInact(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

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

take(0, XS) → nil
zip(nil, XS) → nil
zip(X, nil) → nil
repItems(nil) → nil
take(x1, x2) → takeInact(x1, x2)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = x1   
POL(nil) = 1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(takeInact(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipInact(x1, x2)) = x1 + x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)
a(takeInact(x1, x2)) → take(x1, x2)

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

a(takeInact(x1, x2)) → take(x1, x2)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = x1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + 2·x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = x1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 2 + x1 + 2·x2   
POL(takeInact(x1, x2)) = 2 + x1 + x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipInact(x1, x2)) = x1 + x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

take(s(N), cons(X, XS)) → cons(X, takeInact(N, a(XS)))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = 2·x1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(takeInact(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + 2·x2   
POL(zipInact(x1, x2)) = x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
repItems(x1) → repItemsInact(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

repItems(x1) → repItemsInact(x1)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = x1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(repItemsInact(x1)) = 1 + x1   
POL(s(x1)) = 2·x1   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipInact(x1, x2)) = x1 + x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
zip(x1, x2) → zipInact(x1, x2)
a(zipInact(x1, x2)) → zip(x1, x2)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), zipInact(a(XS), a(YS)))
zip(x1, x2) → zipInact(x1, x2)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = x1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = x1   
POL(s(x1)) = 2·x1   
POL(zip(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zipInact(x1, x2)) = 1 + x1 + x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(zipInact(x1, x2)) → zip(x1, x2)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(zipInact(x1, x2)) → zip(x1, x2)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

a(zipInact(x1, x2)) → zip(x1, x2)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(consInact(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(incrInact(x1)) = x1   
POL(oddNs) = 0   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsInact(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(zip(x1, x2)) = x1 + x2   
POL(zipInact(x1, x2)) = 1 + 2·x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

repItems(cons(X, XS)) → cons(X, consInact(X, repItemsInact(a(XS))))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(consInact(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = x1   
POL(oddNs) = 0   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(repItemsInact(x1)) = 1 + x1   
POL(s(x1)) = x1   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
a(repItemsInact(x1)) → repItems(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

a(repItemsInact(x1)) → repItems(x1)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a(x1)) = x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(consInact(x1, x2)) = x1 + 2·x2   
POL(incr(x1)) = 2·x1   
POL(incrInact(x1)) = 2·x1   
POL(oddNs) = 0   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + x1   
POL(repItemsInact(x1)) = 2 + x1   
POL(s(x1)) = x1   




↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.

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

A(incrInact(x1)) → INCR(x1)
A(consInact(x1, x2)) → CONS(x1, x2)
INCR(cons(X, XS)) → CONS(s(X), incrInact(a(XS)))
PAIRNS → ODDNS
ODDNS → PAIRNS
ODDNS → INCR(pairNs)
INCR(cons(X, XS)) → A(XS)
PAIRNS → CONS(0, incrInact(oddNs))

The TRS R consists of the following rules:

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

A(incrInact(x1)) → INCR(x1)
A(consInact(x1, x2)) → CONS(x1, x2)
INCR(cons(X, XS)) → CONS(s(X), incrInact(a(XS)))
PAIRNS → ODDNS
ODDNS → PAIRNS
ODDNS → INCR(pairNs)
INCR(cons(X, XS)) → A(XS)
PAIRNS → CONS(0, incrInact(oddNs))

The TRS R consists of the following rules:

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 4 less nodes.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ AND
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                  ↳ QDP
  ↳ Incomplete Giesl Middeldorp-Transformation

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

A(incrInact(x1)) → INCR(x1)
INCR(cons(X, XS)) → A(XS)

The TRS R consists of the following rules:

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ AND
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ QDPSizeChangeProof
                                                  ↳ QDP
  ↳ Incomplete Giesl Middeldorp-Transformation

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

A(incrInact(x1)) → INCR(x1)
INCR(cons(X, XS)) → A(XS)

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: 



↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ AND
                                                  ↳ QDP
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

PAIRNS → ODDNS
ODDNS → PAIRNS

The TRS R consists of the following rules:

pairNs → cons(0, incrInact(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)
cons(x1, x2) → consInact(x1, x2)
a(consInact(x1, x2)) → cons(x1, x2)

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

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ RRRPoloQTRSProof
                                ↳ QTRS
                                  ↳ RRRPoloQTRSProof
                                    ↳ QTRS
                                      ↳ RRRPoloQTRSProof
                                        ↳ QTRS
                                          ↳ DependencyPairsProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ AND
                                                  ↳ QDP
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ NonTerminationProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

PAIRNS → ODDNS
ODDNS → PAIRNS

R is empty.
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:

PAIRNS → ODDNS
ODDNS → PAIRNS

The TRS R consists of the following rules:none


s = ODDNS evaluates to t =ODDNS

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



Rewriting sequence

ODDNS → PAIRNS
with rule ODDNS → PAIRNS at position [] and matcher [ ]

PAIRNS → ODDNS
with rule PAIRNS → ODDNS

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.




We applied the Incomplete Giesl Middeldorp [11] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
tailActive(cons(X, XS)) → mark(XS)
repItemsActive(nil) → nil
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

Q is empty.

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
tailActive(cons(X, XS)) → mark(XS)
repItemsActive(nil) → nil
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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

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

POL(0) = 0   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 1 + x1   
POL(tailActive(x1)) = 1 + x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(takeActive(x1, x2)) = 2·x1 + x2   
POL(zip(x1, x2)) = x1 + 2·x2   
POL(zipActive(x1, x2)) = x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(nil) → nil
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

Q is empty.

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(nil) → nil
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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

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

POL(0) = 0   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = x1 + 2·x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(repItemsActive(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(tailActive(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(takeActive(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipActive(x1, x2)) = 2·x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

Q is empty.

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(0, XS) → nil
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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

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

POL(0) = 0   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(takeActive(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = x1 + x2   
POL(zipActive(x1, x2)) = x1 + x2   




↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

Q is empty.

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(nil, XS) → nil
zipActive(X, nil) → nil
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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

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

POL(0) = 0   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = 2·x1 + 2·x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(takeActive(x1, x2)) = x1 + 2·x2   
POL(zip(x1, x2)) = 2·x1 + 2·x2   
POL(zipActive(x1, x2)) = 2·x1 + 2·x2   




↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof

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

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
MARK(zip(x1, x2)) → MARK(x2)
REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(pairNs) → PAIRNSACTIVE
MARK(zip(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(take(x1, x2)) → MARK(x1)
MARK(tail(x1)) → TAILACTIVE(mark(x1))
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(zip(x1, x2)) → ZIPACTIVE(mark(x1), mark(x2))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(tail(x1)) → MARK(x1)
MARK(repItems(x1)) → MARK(x1)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))
ODDNSACTIVE → PAIRNSACTIVE

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ 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)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
MARK(zip(x1, x2)) → MARK(x2)
REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(pairNs) → PAIRNSACTIVE
MARK(zip(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(take(x1, x2)) → MARK(x1)
MARK(tail(x1)) → TAILACTIVE(mark(x1))
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(zip(x1, x2)) → ZIPACTIVE(mark(x1), mark(x2))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(tail(x1)) → MARK(x1)
MARK(repItems(x1)) → MARK(x1)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))
ODDNSACTIVE → PAIRNSACTIVE

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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.

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ 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)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(zip(x1, x2)) → MARK(x2)
MARK(pair(x1, x2)) → MARK(x2)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(Y)
MARK(take(x1, x2)) → MARK(x1)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(zip(x1, x2)) → ZIPACTIVE(mark(x1), mark(x2))
MARK(zip(x1, x2)) → MARK(x1)
MARK(s(x1)) → MARK(x1)
MARK(tail(x1)) → MARK(x1)
MARK(repItems(x1)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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(x2)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(Y)
ZIPACTIVE(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(x1, x2)) → ZIPACTIVE(mark(x1), mark(x2))
MARK(zip(x1, x2)) → MARK(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(ODDNSACTIVE) = 0   
POL(REPITEMSACTIVE(x1)) = 2·x1   
POL(TAKEACTIVE(x1, x2)) = 2·x1 + 2·x2   
POL(ZIPACTIVE(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = 2·x1 + 2·x2   
POL(takeActive(x1, x2)) = 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   
POL(zipActive(x1, x2)) = 2 + 2·x1 + x2   



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ 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)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(pair(x1, x2)) → MARK(x2)
MARK(take(x1, x2)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(tail(x1)) → MARK(x1)
MARK(repItems(x1)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(incr(x1)) → MARK(x1)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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(x1)) → MARK(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = 2·x1   
POL(MARK(x1)) = x1   
POL(ODDNSACTIVE) = 0   
POL(REPITEMSACTIVE(x1)) = x1   
POL(TAKEACTIVE(x1, x2)) = x1 + 2·x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 1 + x1   
POL(tailActive(x1)) = 1 + x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(takeActive(x1, x2)) = x1 + 2·x2   
POL(zip(x1, x2)) = x1 + x2   
POL(zipActive(x1, x2)) = x1 + x2   



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ 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)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(pair(x1, x2)) → MARK(x2)
MARK(take(x1, x2)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(repItems(x1)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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:

REPITEMSACTIVE(cons(X, XS)) → MARK(X)
MARK(repItems(x1)) → MARK(x1)
MARK(repItems(x1)) → REPITEMSACTIVE(mark(x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = 2·x1   
POL(MARK(x1)) = 2·x1   
POL(ODDNSACTIVE) = 0   
POL(REPITEMSACTIVE(x1)) = 1 + 2·x1   
POL(TAKEACTIVE(x1, x2)) = 2·x1 + 2·x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(repItemsActive(x1)) = 1 + 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = 2·x1 + x2   
POL(takeActive(x1, x2)) = 2·x1 + x2   
POL(zip(x1, x2)) = 2·x1 + x2   
POL(zipActive(x1, x2)) = 2·x1 + x2   



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof

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

MARK(take(x1, x2)) → MARK(x2)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
MARK(pair(x1, x2)) → MARK(x2)
MARK(take(x1, x2)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(oddNs) → ODDNSACTIVE
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
TAKEACTIVE(s(N), cons(X, XS)) → MARK(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = 2·x1   
POL(MARK(x1)) = 2·x1   
POL(ODDNSACTIVE) = 0   
POL(TAKEACTIVE(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 2   
POL(oddNs) = 0   
POL(oddNsActive) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(pairNsActive) = 0   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(takeActive(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = x1 + x2   
POL(zipActive(x1, x2)) = x1 + x2   



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ RuleRemovalProof

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

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(pair(x1, x2)) → MARK(x1)
ODDNSACTIVE → INCRACTIVE(pairNsActive)
MARK(s(x1)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
MARK(oddNs) → ODDNSACTIVE
MARK(incr(x1)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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:

ODDNSACTIVE → INCRACTIVE(pairNsActive)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = x1   
POL(MARK(x1)) = 2·x1   
POL(ODDNSACTIVE) = 2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 1   
POL(oddNsActive) = 1   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 1   
POL(pairNsActive) = 1   
POL(repItems(x1)) = 2·x1   
POL(repItemsActive(x1)) = 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(tailActive(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(takeActive(x1, x2)) = x1 + x2   
POL(zip(x1, x2)) = x1 + x2   
POL(zipActive(x1, x2)) = x1 + x2   



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ RuleRemovalProof
                                                ↳ QDP
                                                  ↳ DependencyGraphProof

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

MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
MARK(oddNs) → ODDNSACTIVE
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(incr(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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.

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ RuleRemovalProof
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ QDPOrderProof

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

MARK(pair(x1, x2)) → MARK(x1)
INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
MARK(incr(x1)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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(pair(x1, x2)) → MARK(x1)
MARK(pair(x1, x2)) → MARK(x2)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(cons(x1, x2)) → MARK(x1)
The remaining pairs can at least be oriented weakly.

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(incr(x1)) → MARK(x1)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = x1   
POL(MARK(x1)) = 1 + x1   
POL(cons(x1, x2)) = 1 + x1   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 1   
POL(oddNsActive) = 1   
POL(pair(x1, x2)) = 1 + x1 + x2   
POL(pairNs) = 1   
POL(pairNsActive) = 1   
POL(repItems(x1)) = x1   
POL(repItemsActive(x1)) = x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 0   
POL(tailActive(x1)) = 0   
POL(take(x1, x2)) = x2   
POL(takeActive(x1, x2)) = x2   
POL(zip(x1, x2)) = x1 + x2   
POL(zipActive(x1, x2)) = x1 + x2   

The following usable rules [17] were oriented:

oddNsActive → oddNs
mark(nil) → nil
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(incr(x1)) → incrActive(mark(x1))
zipActive(x1, x2) → zip(x1, x2)
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
mark(repItems(x1)) → repItemsActive(mark(x1))
pairNsActive → pairNs
mark(0) → 0
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
mark(cons(x1, x2)) → cons(mark(x1), x2)
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
repItemsActive(x1) → repItems(x1)
mark(oddNs) → oddNsActive
mark(s(x1)) → s(mark(x1))
incrActive(x1) → incr(x1)
oddNsActive → incrActive(pairNsActive)
mark(tail(x1)) → tailActive(mark(x1))
mark(pairNs) → pairNsActive
pairNsActive → cons(0, incr(oddNs))
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
tailActive(x1) → tail(x1)
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ RuleRemovalProof
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ QDPOrderProof
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof

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

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(incr(x1)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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.

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ QDP
                                              ↳ RuleRemovalProof
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ QDPOrderProof
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
                                                            ↳ QDP
                                                              ↳ UsableRulesProof

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

MARK(s(x1)) → MARK(x1)
MARK(incr(x1)) → MARK(x1)

The TRS R consists of the following rules:

mark(pairNs) → pairNsActive
pairNsActive → pairNs
mark(oddNs) → oddNsActive
oddNsActive → oddNs
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(zip(x1, x2)) → zipActive(mark(x1), mark(x2))
zipActive(x1, x2) → zip(x1, x2)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(repItems(x1)) → repItemsActive(mark(x1))
repItemsActive(x1) → repItems(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
mark(pair(x1, x2)) → pair(mark(x1), mark(x2))
pairNsActive → cons(0, incr(oddNs))
oddNsActive → incrActive(pairNsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
takeActive(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
zipActive(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
repItemsActive(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))

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.

↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ RuleRemovalProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ RuleRemovalProof
                                            ↳ 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(x1)) → MARK(x1)
MARK(incr(x1)) → 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: