Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

NATS → NATS
PAIRS → ODDS
ODDS → INCR(pairs)
ODDS → PAIRS
INCR(cons(X, XS)) → ACTIVATE(XS)
TAIL(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__incr(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__incr(x1)  =  n__incr(x1)
INCR(x1)  =  x1
cons(x1, x2)  =  x2

Recursive path order with status [RPO].
Precedence:

trivial

Status:

n_incr1: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

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

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

(8) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(9) TRUE

(10) Obligation:

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

ODDS → PAIRS
PAIRS → ODDS

The TRS R consists of the following rules:

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

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

(11) Obligation:

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

NATS → NATS

The TRS R consists of the following rules:

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

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