Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(odds) -> A__ODDS
MARK1(head1(X)) -> A__HEAD1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(pairs) -> A__PAIRS
MARK1(head1(X)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(nats) -> A__NATS
MARK1(tail1(X)) -> MARK1(X)
A__ODDS -> A__PAIRS
A__HEAD1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(odds) -> A__ODDS
MARK1(head1(X)) -> A__HEAD1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(pairs) -> A__PAIRS
MARK1(head1(X)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(nats) -> A__NATS
MARK1(tail1(X)) -> MARK1(X)
A__ODDS -> A__PAIRS
A__HEAD1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(odds) -> A__ODDS
MARK1(head1(X)) -> A__HEAD1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(head1(X)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(tail1(X)) -> MARK1(X)
A__HEAD1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(head1(X)) -> A__HEAD1(mark1(X))
MARK1(head1(X)) -> MARK1(X)
MARK1(tail1(X)) -> MARK1(X)
A__HEAD1(cons2(X, XS)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(odds) -> A__ODDS
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__HEAD1(x1)) = 1 + x1   
POL(A__INCR1(x1)) = 2·x1   
POL(A__ODDS) = 0   
POL(A__TAIL1(x1)) = 2 + 2·x1   
POL(MARK1(x1)) = 2·x1   
POL(a__head1(x1)) = 3 + 2·x1   
POL(a__incr1(x1)) = 2·x1   
POL(a__nats) = 0   
POL(a__odds) = 0   
POL(a__pairs) = 0   
POL(a__tail1(x1)) = 3 + x1   
POL(cons2(x1, x2)) = 2·x1 + 2·x2   
POL(head1(x1)) = 3 + 2·x1   
POL(incr1(x1)) = 2·x1   
POL(mark1(x1)) = x1   
POL(nats) = 0   
POL(nil) = 1   
POL(odds) = 0   
POL(pairs) = 0   
POL(s1(x1)) = x1   
POL(tail1(x1)) = 3 + x1   

The following usable rules [14] were oriented:

a__incr1(X) -> incr1(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__odds -> a__incr1(a__pairs)
mark1(pairs) -> a__pairs
a__tail1(X) -> tail1(X)
a__head1(cons2(X, XS)) -> mark1(X)
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nil) -> nil
a__odds -> odds
a__pairs -> cons2(0, incr1(odds))
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
a__nats -> cons2(0, incr1(nats))
a__pairs -> pairs
mark1(0) -> 0
a__head1(X) -> head1(X)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__nats -> nats
mark1(nats) -> a__nats



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(odds) -> A__ODDS
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(odds) -> A__ODDS
The remaining pairs can at least be oriented weakly.

MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__INCR1(x1)) = x1   
POL(A__ODDS) = 1   
POL(MARK1(x1)) = 2·x1   
POL(a__head1(x1)) = x1   
POL(a__incr1(x1)) = x1   
POL(a__nats) = 0   
POL(a__odds) = 1   
POL(a__pairs) = 1   
POL(a__tail1(x1)) = 1 + x1   
POL(cons2(x1, x2)) = 2·x1 + x2   
POL(head1(x1)) = x1   
POL(incr1(x1)) = x1   
POL(mark1(x1)) = x1   
POL(nats) = 0   
POL(nil) = 0   
POL(odds) = 1   
POL(pairs) = 1   
POL(s1(x1)) = x1   
POL(tail1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

a__incr1(X) -> incr1(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__odds -> a__incr1(a__pairs)
mark1(pairs) -> a__pairs
a__tail1(X) -> tail1(X)
a__head1(cons2(X, XS)) -> mark1(X)
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nil) -> nil
a__odds -> odds
a__pairs -> cons2(0, incr1(odds))
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
a__nats -> cons2(0, incr1(nats))
mark1(0) -> 0
a__pairs -> pairs
a__head1(X) -> head1(X)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__nats -> nats
mark1(nats) -> a__nats



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

MARK1(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP

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

MARK1(s1(X)) -> MARK1(X)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
A__INCR1(cons2(X, XS)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__nats -> cons2(0, incr1(nats))
a__pairs -> cons2(0, incr1(odds))
a__odds -> a__incr1(a__pairs)
a__incr1(cons2(X, XS)) -> cons2(s1(mark1(X)), incr1(XS))
a__head1(cons2(X, XS)) -> mark1(X)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(nats) -> a__nats
mark1(pairs) -> a__pairs
mark1(odds) -> a__odds
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(head1(X)) -> a__head1(mark1(X))
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__nats -> nats
a__pairs -> pairs
a__odds -> odds
a__incr1(X) -> incr1(X)
a__head1(X) -> head1(X)
a__tail1(X) -> tail1(X)

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