Termination w.r.t. Q proof of TRS/CSREx1_Luc04b

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

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.