Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__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)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + 2·x₂   
POL(head(x₁)) = 2 + 2·x₁   
POL(incr(x₁)) = x₁   
POL(n__incr(x₁)) = x₁   
POL(n__nats) = 0   
POL(n__odds) = 0   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x₁)) = x₁   
POL(tail(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

head(cons(X, XS)) → X

(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(incr(x₁)) = x₁   
POL(n__incr(x₁)) = x₁   
POL(n__nats) = 0   
POL(n__odds) = 0   
POL(nats) = 0   
POL(odds) = 0   
POL(pairs) = 0   
POL(s(x₁)) = x₁   
POL(tail(x₁)) = 2 + 2·x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

tail(cons(X, XS)) → activate(XS)

(4) Obligation:

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

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

ODDS → INCR(pairs)
ODDS → PAIRS
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__nats) → NATS
ACTIVATE(n__odds) → ODDS

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__odds) → ODDS
ODDS → INCR(pairs)

The TRS R consists of the following rules:

nats → cons(0, n__incr(n__nats))
pairs → cons(0, n__incr(n__odds))
odds → incr(pairs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
incr(X) → n__incr(X)
nats → n__nats
odds → n__odds
activate(n__incr(X)) → incr(activate(X))
activate(n__nats) → nats
activate(n__odds) → odds
activate(X) → X

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

(9) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / 0] on the rule
INCR(cons(0, n__incr(n__nats)))[ ]ⁿ[ ] → INCR(cons(0, n__incr(n__nats)))[ ]ⁿ[x0 / 0]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu
INCR(cons(x0, n__incr(n__nats)))[ ]ⁿ[ ] → INCR(cons(0, n__incr(n__nats)))[ ]ⁿ[ ]
    by Narrowing at position: [0]
        intermediate steps: Instantiation
        INCR(cons(x1, n__incr(n__nats)))[ ]ⁿ[ ] → INCR(nats)[ ]ⁿ[ ]
            by Narrowing at position: [0]
                intermediate steps: Instantiation - Instantiation
                INCR(cons(x1, n__incr(y0)))[ ]ⁿ[ ] → INCR(activate(y0))[ ]ⁿ[ ]
                    by Narrowing at position: []
                        intermediate steps: Instantiation - Instantiation
                        INCR(cons(X, XS))[ ]ⁿ[ ] → ACTIVATE(XS)[ ]ⁿ[ ]
                            by OriginalRule from TRS P

                        intermediate steps: Instantiation
                        ACTIVATE(n__incr(X))[ ]ⁿ[ ] → INCR(activate(X))[ ]ⁿ[ ]
                            by OriginalRule from TRS P

                activate(n__nats)[ ]ⁿ[ ] → nats[ ]ⁿ[ ]
                    by OriginalRule from TRS R

        nats[ ]ⁿ[ ] → cons(0, n__incr(n__nats))[ ]ⁿ[ ]
            by OriginalRule from TRS R

(10) NO