Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

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

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

POL(0) = 0   
POL(a__head(x1)) = 2·x1   
POL(a__incr(x1)) = 2·x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(a__tail(x1)) = 2 + x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(head(x1)) = 2·x1   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   
POL(nats) = 0   
POL(nil) = 2   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

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

mark(tail(X)) → a__tail(mark(X))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__head(x1)) = x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(a__tail(x1)) = 1 + 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(head(x1)) = x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2 + 2·x1   




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

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

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

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

POL(0) = 0   
POL(a__head(x1)) = 2 + x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(head(x1)) = 2 + x1   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   




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

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)

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

a__head(X) → head(X)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__head(x1)) = 2 + 2·x1   
POL(a__incr(x1)) = 2·x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(head(x1)) = 1 + 2·x1   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   




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

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

mark(head(X)) → a__head(mark(X))
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__head(x1)) = 2·x1   
POL(a__incr(x1)) = 2·x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(head(x1)) = 1 + 2·x1   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   




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

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

a__nats → nats
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(a__incr(x1)) = 2·x1   
POL(a__nats) = 2   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   
POL(nats) = 1   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x1)) = x1   




↳ 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:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.

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

MARK(pairs) → A__PAIRS
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__ODDS → A__PAIRS
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → A__NATS
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)
A__ODDS → A__INCR(a__pairs)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

↳ 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:

MARK(pairs) → A__PAIRS
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__ODDS → A__PAIRS
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → A__NATS
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)
A__ODDS → A__INCR(a__pairs)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

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

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ODDS → A__INCR(a__pairs)
MARK(odds) → A__ODDS
A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(odds) → A__ODDS


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(A__INCR(x1)) = 1 + x1   
POL(A__ODDS) = 2   
POL(MARK(x1)) = 1 + 2·x1   
POL(a__incr(x1)) = x1   
POL(a__nats) = 0   
POL(a__odds) = 1   
POL(a__pairs) = 1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nats) = 0   
POL(odds) = 1   
POL(pairs) = 1   
POL(s(x1)) = x1   



↳ 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:

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

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

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

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

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

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

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__INCR(cons(X, XS)) → MARK(X)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( cons(x1, x2) ) =
/1\
\0/
+
/10\
\10/
·x1+
/00\
\00/
·x2

M( odds ) =
/1\
\0/

M( a__pairs ) =
/1\
\0/

M( a__nats ) =
/1\
\1/

M( a__incr(x1) ) =
/0\
\0/
+
/11\
\01/
·x1

M( incr(x1) ) =
/0\
\0/
+
/11\
\01/
·x1

M( a__odds ) =
/1\
\0/

M( pairs ) =
/1\
\0/

M( mark(x1) ) =
/0\
\0/
+
/10\
\01/
·x1

M( s(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

M( nats ) =
/1\
\1/

M( 0 ) =
/0\
\0/

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

M( A__INCR(x1) ) = 0+
[0,1]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__odds → odds
a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))



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

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

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

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

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

POL(0) = 0   
POL(A__INCR(x1)) = x1   
POL(MARK(x1)) = x1   
POL(a__incr(x1)) = 1 + x1   
POL(a__nats) = 1   
POL(a__odds) = 1   
POL(a__pairs) = 0   
POL(cons(x1, x2)) = x1   
POL(incr(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nats) = 1   
POL(odds) = 1   
POL(pairs) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

a__pairs → pairs
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__odds → odds
a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))



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

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

A__INCR(cons(X, XS)) → MARK(X)

The TRS R consists of the following rules:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.