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(head1(X)) -> A__HEAD1(mark1(X))
MARK1(head1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

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
A__INCR1(cons2(X, XS)) -> 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)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( pairs ) = max{0, -1}


POL( a__pairs ) = max{0, -3}


POL( odds ) = max{0, -3}


POL( mark1(x1) ) = 2x1


POL( A__HEAD1(x1) ) = x1


POL( MARK1(x1) ) = x1


POL( 0 ) = 0


POL( A__TAIL1(x1) ) = x1


POL( a__nats ) = max{0, -3}


POL( head1(x1) ) = 2x1 + 2


POL( a__tail1(x1) ) = 3x1


POL( a__odds ) = max{0, -3}


POL( nil ) = 3


POL( cons2(x1, x2) ) = x1 + 2x2


POL( A__INCR1(x1) ) = x1


POL( incr1(x1) ) = 2x1


POL( A__ODDS ) = max{0, -3}


POL( nats ) = 0


POL( a__incr1(x1) ) = 2x1


POL( s1(x1) ) = x1


POL( a__head1(x1) ) = 2x1 + 2


POL( tail1(x1) ) = 3x1



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(tail1(X)) -> MARK1(X)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
A__HEAD1(cons2(X, XS)) -> 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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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(s1(X)) -> MARK1(X)
A__ODDS -> A__INCR1(a__pairs)
MARK1(tail1(X)) -> MARK1(X)
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(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(tail1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

A__TAIL1(cons2(X, XS)) -> MARK1(XS)
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)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( pairs ) = max{0, -3}


POL( a__pairs ) = max{0, -3}


POL( odds ) = max{0, -3}


POL( mark1(x1) ) = x1


POL( MARK1(x1) ) = x1 + 1


POL( a__nats ) = max{0, -3}


POL( 0 ) = max{0, -3}


POL( A__TAIL1(x1) ) = 2x1 + 1


POL( head1(x1) ) = x1 + 1


POL( a__tail1(x1) ) = 2x1 + 1


POL( a__odds ) = max{0, -3}


POL( nil ) = 2


POL( cons2(x1, x2) ) = x1 + 2x2


POL( A__INCR1(x1) ) = 2x1 + 1


POL( incr1(x1) ) = 2x1


POL( A__ODDS ) = 1


POL( nats ) = max{0, -2}


POL( a__incr1(x1) ) = 2x1


POL( s1(x1) ) = 2x1


POL( a__head1(x1) ) = x1 + 1


POL( tail1(x1) ) = 2x1 + 1



The following usable rules [14] were oriented:

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



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

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

A__TAIL1(cons2(X, XS)) -> MARK1(XS)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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


A__ODDS -> A__INCR1(a__pairs)
MARK1(odds) -> A__ODDS
The remaining pairs can at least be oriented weakly.

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)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( pairs ) = 2


POL( a__pairs ) = 2


POL( odds ) = 2


POL( mark1(x1) ) = x1


POL( MARK1(x1) ) = 2x1


POL( a__nats ) = 1


POL( 0 ) = max{0, -3}


POL( head1(x1) ) = 2x1


POL( a__tail1(x1) ) = x1


POL( a__odds ) = 2


POL( nil ) = 1


POL( cons2(x1, x2) ) = 2x1 + x2


POL( A__INCR1(x1) ) = x1


POL( incr1(x1) ) = x1


POL( A__ODDS ) = 3


POL( nats ) = 1


POL( a__incr1(x1) ) = x1


POL( s1(x1) ) = x1


POL( a__head1(x1) ) = 2x1


POL( tail1(x1) ) = x1



The following usable rules [14] were oriented:

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



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