YES Termination proof of ../tpdb/TRS/CSR/Ex1_Luc04b.trs
Termination of the following Term Rewriting System could be proven:

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

nats → cons(0, incr(nats))
pairs → cons(0, incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incr(XS))
head(cons(X, XS)) → X
tail(cons(X, XS)) → XS

The replacement map contains the following entries:

nats: empty set
cons: {1}
0: empty set
incr: {1}
pairs: empty set
odds: empty set
s: {1}
head: {1}
tail: {1}


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

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

nats → cons(0, incr(nats))
pairs → cons(0, incr(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incr(XS))
head(cons(X, XS)) → X
tail(cons(X, XS)) → XS

The replacement map contains the following entries:

nats: empty set
cons: {1}
0: empty set
incr: {1}
pairs: empty set
odds: empty set
s: {1}
head: {1}
tail: {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:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → a(XS)
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

Q is empty.

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

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
head(cons(X, XS)) → X
tail(cons(X, XS)) → a(XS)
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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

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

POL(0) = 0   
POL(a(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(head(x1)) = 1 + x1   
POL(incr(x1)) = x1   
POL(incrInact(x1)) = x1   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   




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

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

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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:

NATS → NATS
A(incrInact(x1)) → INCR(x1)
PAIRS → ODDS
ODDS → PAIRS
INCR(cons(X, XS)) → A(XS)
ODDS → INCR(pairs)

The TRS R consists of the following rules:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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

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

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

NATS → NATS
A(incrInact(x1)) → INCR(x1)
PAIRS → ODDS
ODDS → PAIRS
INCR(cons(X, XS)) → A(XS)
ODDS → INCR(pairs)

The TRS R consists of the following rules:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 1 less node.

↳ CSR
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ 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:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
                  ↳ QDP
                  ↳ 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
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                  ↳ QDP
  ↳ Incomplete Giesl Middeldorp-Transformation

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

PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ NonTerminationProof
                  ↳ QDP
  ↳ Incomplete Giesl Middeldorp-Transformation

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

PAIRS → ODDS
ODDS → PAIRS

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:

PAIRS → ODDS
ODDS → PAIRS

The TRS R consists of the following rules:none


s = ODDS evaluates to t =ODDS

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



Rewriting sequence

ODDS → PAIRS
with rule ODDS → PAIRS at position [] and matcher [ ]

PAIRS → ODDS
with rule PAIRS → ODDS

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.





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

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

NATS → NATS

The TRS R consists of the following rules:

nats → cons(0, incrInact(nats))
pairs → cons(0, incrInact(odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), incrInact(a(XS)))
a(x) → x
incr(x1) → incrInact(x1)
a(incrInact(x1)) → incr(x1)

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
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ NonTerminationProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

NATS → NATS

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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

NATS → NATS

The TRS R consists of the following rules:none


s = NATS evaluates to t =NATS

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS to NATS.




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(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(head(x1)) → headActive(mark(x1))
headActive(x1) → head(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
headActive(cons(X, XS)) → mark(X)
tailActive(cons(X, XS)) → mark(XS)

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(head(x1)) → headActive(mark(x1))
headActive(x1) → head(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
headActive(cons(X, XS)) → mark(X)
tailActive(cons(X, XS)) → mark(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:

mark(head(x1)) → headActive(mark(x1))
headActive(cons(X, XS)) → mark(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons(x1, x2)) = x1 + x2   
POL(head(x1)) = 2 + 2·x1   
POL(headActive(x1)) = 2 + 2·x1   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(tailActive(x1)) = 2·x1   




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

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
headActive(x1) → head(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
tailActive(cons(X, XS)) → mark(XS)

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
headActive(x1) → head(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
tailActive(cons(X, XS)) → mark(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:

headActive(x1) → head(x1)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(head(x1)) = 1 + x1   
POL(headActive(x1)) = 2 + x1   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 2·x1   
POL(tailActive(x1)) = 2·x1   




↳ 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(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
tailActive(cons(X, XS)) → mark(XS)

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
tailActive(cons(X, XS)) → mark(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)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 1 + 2·x1   
POL(tailActive(x1)) = 1 + 2·x1   




↳ 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(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(tail(x1)) → tailActive(mark(x1))
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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:

mark(tail(x1)) → tailActive(mark(x1))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(incr(x1)) = 2·x1   
POL(incrActive(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + 2·x1   
POL(tailActive(x1)) = 2 + 2·x1   




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

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
tailActive(x1) → tail(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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(x1) → tail(x1)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + 2·x1   
POL(tailActive(x1)) = 2 + 2·x1   




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

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))

Q is empty.

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

mark(nats) → natsActive
natsActive → nats
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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:

natsActive → nats
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)) = 2·x1   
POL(nats) = 1   
POL(natsActive) = 2   
POL(odds) = 0   
POL(oddsActive) = 0   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = x1   




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

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

mark(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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:

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(pairs) → PAIRSACTIVE
MARK(s(x1)) → MARK(x1)
ODDSACTIVE → PAIRSACTIVE
MARK(odds) → ODDSACTIVE
MARK(nats) → NATSACTIVE
ODDSACTIVE → INCRACTIVE(pairsActive)
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(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof

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

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(pairs) → PAIRSACTIVE
MARK(s(x1)) → MARK(x1)
ODDSACTIVE → PAIRSACTIVE
MARK(odds) → ODDSACTIVE
MARK(nats) → NATSACTIVE
ODDSACTIVE → INCRACTIVE(pairsActive)
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(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof

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

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
MARK(odds) → ODDSACTIVE
ODDSACTIVE → INCRACTIVE(pairsActive)
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(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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(odds) → ODDSACTIVE


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = x1   
POL(MARK(x1)) = 2·x1   
POL(ODDSACTIVE) = 1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(incrActive(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 1   
POL(oddsActive) = 1   
POL(pairs) = 1   
POL(pairsActive) = 1   
POL(s(x1)) = x1   



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

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

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(s(x1)) → MARK(x1)
ODDSACTIVE → INCRACTIVE(pairsActive)
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(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ QDPOrderProof

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)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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(s(x1)) → MARK(x1)
MARK(incr(x1)) → MARK(x1)
MARK(incr(x1)) → INCRACTIVE(mark(x1))
The remaining pairs can at least be oriented weakly.

INCRACTIVE(cons(X, XS)) → MARK(X)
MARK(cons(x1, x2)) → MARK(x1)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(INCRACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(cons(x1, x2)) = x1   
POL(incr(x1)) = 1 + x1   
POL(incrActive(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(natsActive) = 0   
POL(odds) = 1   
POL(oddsActive) = 1   
POL(pairs) = 0   
POL(pairsActive) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

mark(nats) → natsActive
mark(cons(x1, x2)) → cons(mark(x1), x2)
oddsActive → incrActive(pairsActive)
mark(odds) → oddsActive
mark(0) → 0
pairsActive → cons(0, incr(odds))
oddsActive → odds
natsActive → cons(0, incr(nats))
mark(s(x1)) → s(mark(x1))
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(pairs) → pairsActive
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(XS))
pairsActive → pairs



↳ CSR
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RRRPoloQTRSProof
                ↳ QTRS
                  ↳ RRRPoloQTRSProof
                    ↳ QTRS
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ 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(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ QDPOrderProof
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ UsableRulesProof

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

MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(nats) → natsActive
mark(pairs) → pairsActive
pairsActive → pairs
mark(odds) → oddsActive
oddsActive → odds
mark(incr(x1)) → incrActive(mark(x1))
incrActive(x1) → incr(x1)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(s(x1)) → s(mark(x1))
natsActive → cons(0, incr(nats))
pairsActive → cons(0, incr(odds))
oddsActive → incrActive(pairsActive)
incrActive(cons(X, XS)) → cons(s(mark(X)), incr(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
                      ↳ RRRPoloQTRSProof
                        ↳ QTRS
                          ↳ RRRPoloQTRSProof
                            ↳ QTRS
                              ↳ DependencyPairsProof
                                ↳ 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(cons(x1, x2)) → MARK(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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