**Proof.** Assume the contrary that *ω* satisfies case (D1) or (D3).

Assume first that *ω* satisfies case (D3) for *u* ≥ *u*<sup>1</sup> ≥ *u*0. From Lemma 4, we obtain that (7) holds. Integrating (7) from *u*<sup>1</sup> to *u*, we find

$$-a(u)\omega^{(n-1)}(u) \quad \geq \ -a(u\_1)\omega^{(n-1)}(u\_1) + \int\_{u\_1}^{u} \mathcal{Q}(\upsilon)F(\omega(\mathcal{g}(\upsilon)))d\upsilon$$

$$\geq \ \int\_{u\_1}^{u} \mathcal{Q}(\upsilon)F(\omega(\mathcal{g}(\upsilon)))d\upsilon. \tag{14}$$

Next, it follows from Lemma 2 that (5) holds. From (5) with *s* = 0, we note that the function

$$
\omega + \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d} \mu^{n-1}} \omega \right) \cdot \mathcal{E}\_{n-2},
$$

or

is positive for *u* ≥ *u*1. Then, from (5) with *s* = 1,

$$\begin{aligned} \frac{\mathbf{d}}{\mathbf{d}\mu} \Big(\omega + a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-2}\big) &= \quad \omega' - a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-3} + \Big(a \cdot \omega^{(n-1)}\Big)' \cdot \mathcal{E}\_{n-2}\big) \\ &\leq \quad \Big(a \cdot \omega^{(n-1)}\Big)' \cdot \mathcal{E}\_{n-2}\big.\end{aligned}$$

which, with (7), gives

$$\frac{\mathbf{d}}{\mathbf{d}\mu} \left(\omega + a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-2}\right) \le -\mathcal{E}\_{n-2} \cdot \mathcal{Q} \cdot \left[F \circ \omega \circ \mathcal{g}\right] \le 0.$$

Integrating this inequality from *u* to ∞, we obtain

$$
\omega(u) + a(u)\omega^{(n-1)}(u)\mathcal{E}\_{n-2}(u) \ge \int\_u^\infty \mathcal{E}\_{n-2}(v)\mathcal{Q}(v)F(\omega(\mathcal{g}(v)))dv.\tag{15}
$$

Combining (14) and (15), we find

$$\begin{aligned} \omega(u) &\geq -a(u)\omega^{(n-1)}(u)\mathcal{E}\_{n-2}(u) + \int\_{u}^{\infty} \mathcal{E}\_{n-2}(v)\mathcal{Q}(v)F(\omega(\mathcal{g}(v)))d\upsilon \\ &\geq \mathcal{E}\_{n-2}(u)\int\_{u\_1}^{u} \mathcal{Q}(v)F(\omega(\mathcal{g}(v)))d\upsilon + \int\_{u}^{\infty} \mathcal{E}\_{n-2}(v)\mathcal{Q}(v)F(\omega(\mathcal{g}(v)))d\upsilon, \end{aligned}$$

and hence,

$$
\omega(\mathcal{g}(u)) \ge \mathcal{E}\_{n-2}(\mathcal{g}(u)) \int\_{u\_1}^{\mathcal{g}(u)} \mathcal{Q}(v) \mathcal{F}(\omega(\mathcal{g}(v))) \, \mathrm{d}v + \int\_{\mathcal{g}(u)}^u \mathcal{E}\_{n-2}(v) \mathcal{Q}(v) \mathcal{F}(\omega(\mathcal{g}(v))) \, \mathrm{d}v
$$

$$
+ \int\_u^\infty \mathcal{E}\_{n-2}(v) \mathcal{Q}(v) \mathcal{F}(\omega(\mathcal{g}(v))) \, \mathrm{d}v. \tag{16}
$$

Using Lemma 3, we obtain that *ω*/E*n*−<sup>2</sup> is increasing, and so

$$F(\omega(\mathfrak{g}(\upsilon))) \ge F(\mathcal{E}\_{n-2}^{-1}(\mathfrak{g}(\mathfrak{u}))) F(\mathcal{E}\_{n-2}(\mathfrak{g}(\upsilon))) F(\omega(\mathfrak{g}(\mathfrak{u}))), \text{ for } \mathfrak{u} \le \upsilon. \tag{17}$$

Thus, (16) reduces to

$$\begin{split} \frac{\omega(\mathcal{g}(u))}{F(\omega(\mathcal{g}(u)))} &\geq \mathcal{E}\_{n-2}(\mathcal{g}(u)) \int\_{u\_1}^{\mathcal{g}(u)} \mathcal{Q}(v) \mathrm{d}v + \int\_{\mathcal{g}(u)}^u \mathcal{E}\_{n-2}(v) \, \mathcal{Q}(v) \mathrm{d}v \\ &+ F\Big(\mathcal{E}\_{n-2}^{-1}(\mathcal{g}(u))\Big) \int\_u^\infty \mathcal{E}\_{n-2}(v) \, \mathcal{Q}(v) F(\mathcal{E}\_{n-2}(\mathcal{g}(v))) \mathrm{d}v, \end{split} \tag{18}$$

which contradicts (13).

Suppose that *ω* satisfies case (D1) for *u* ≥ *u*<sup>1</sup> ≥ *u*0. From (13), we can demonstrate that ∞

$$\int\_{u\_1}^{\infty} \mathcal{E}\_{n-2}(v) \mathcal{Q}(v) \mathrm{d}v = \infty,$$

by following the same procedure as in the proof of Theorem 1 in [10]. Further, it follows from (H1) that

$$\int\_{u\_1}^{\infty} \mathcal{Q}(v) \mathrm{d}v = \infty.$$

Since *ω* is positive and increasing, we get that *x* ≥ (1 − *p*)*ω*, and there is *u*<sup>2</sup> ≥ *u*<sup>1</sup> with *ω*(*g*(*u*)) ≥ *M* for *u* ≥ *u*2. Integrating (4) from *u*<sup>1</sup> to *u*, we find

$$\begin{split} \mathfrak{a}(\mathfrak{u}\_{1})\mathfrak{w}^{(n-1)}(\mathfrak{u}\_{1}) &\geq \int\_{\mathfrak{u}\_{1}}^{\infty} q(\upsilon) F(\mathfrak{x}(\mathfrak{g}(\upsilon))) \mathrm{d}\upsilon \\ &\geq \int\_{\mathfrak{u}\_{1}}^{\infty} q(\upsilon) F(1 - p(\mathfrak{g}(\upsilon))) F(\mathfrak{w}(\mathfrak{g}(\upsilon))) \mathrm{d}\upsilon \\ &\geq \quad F(M) \int\_{\mathfrak{u}\_{1}}^{\infty} q(\upsilon) F(1 - p(\mathfrak{g}(\upsilon))) \mathrm{d}\upsilon. \end{split} \tag{19}$$

From the fact that [E*n*−2◦*h*] <sup>E</sup>*n*−<sup>2</sup> <sup>≥</sup> 1, we find

$$q(\mu)F(1 - p(\mathcal{g}(\mu))) \ge \mathcal{Q}(\mu),$$

which with (19) gives

$$a(\mathfrak{u}\_1)\omega^{(n-1)}(\mathfrak{u}\_1) \ge F(M) \int\_{\mathfrak{u}\_1}^{\infty} \mathbb{Q}(\upsilon) \mathbf{d}\upsilon\_\prime$$

a contradiction. Therefore, *ω* satisfies case (D2), eventually. The proof is complete.

**Theorem 2.** *Assume that F*(*x*) = *x. If conditions (11) and (13) are satisfied, then Equation (4) becomes oscillatory.*

**Proof.** Assuming that there is a non-oscillatory solution to (4) necessarily means that there is a solution *x* of (4), in which *x* ∈ P*s*. From Lemma 1, the derivatives of the function *ω* have three possibilities. It follows from Lemma 6 that *ω* satisfies case (D2), eventually. Following the same approach in Theorem 1, we can prove that if *ω* fulfills case (D2), then we get a conflict with condition (11). The proof is complete.

In addition to the above, we also present the following theorem, which provides a criterion for the oscillation of (4) based on the comparison principle. So, we need to review the following lemma.

**Lemma 7** (Lemma 2.2.3, [33])**.** *Let* ∈ *<sup>C</sup>s*([*u*0, <sup>∞</sup>))*,* (*u*) <sup>&</sup>gt; <sup>0</sup>*,* lim*u*→<sup>∞</sup> (*u*) <sup>=</sup> <sup>0</sup>*,* (*s*)(*u*) *be of constant sign eventually, and* (*s*) <sup>=</sup> <sup>0</sup> *on a subray of* [*u*0, <sup>∞</sup>)*. If* (*s*−<sup>1</sup>)(*u*)(*s*)(*u*) <sup>&</sup>lt; <sup>0</sup> *for u* ≥ *u*1*, then there is a u<sup>μ</sup>* ≥ *u*<sup>1</sup> *such that*

$$\mathbb{G}(\mu) \ge \frac{\mu}{(s-1)!} \mu^{s-1} \Big| \mathbb{G}^{(s-1)}(\mu) \Big| .$$

*for* <sup>0</sup> <sup>&</sup>lt; *<sup>μ</sup>* <sup>&</sup>lt; <sup>1</sup> *and u* <sup>∈</sup> *uμ*, ∞ *.*

**Theorem 3.** *Assume that* lim*x*→<sup>0</sup> *<sup>x</sup> <sup>F</sup>*(*x*) = *K* < ∞*, and let (8) and (13) hold. The oscillation of the DDE*

$$\psi'(u) + F(\psi(g(u))) \frac{1}{a(u)} \int\_{u\_1}^{u} q(\upsilon) F(1 - p(g(\upsilon))) F\left(\frac{\mu g^{n-2}(\upsilon)}{(n-2)!}\right) d\upsilon = 0,\tag{20}$$

*for some μ* ∈ (0, 1)*, ensures the oscillation of Equation (4).*

**Proof.** Assuming that there is a non-oscillatory solution to (4) necessarily means that there is a solution *x* of (4), in which *x* ∈ P*s*. From Lemma 1, the derivatives of the function *ω* have three possibilities. It follows from Lemma 6 that *ω* satisfies case (D2), eventually.

Using Lemma 7 with = *ω* and *s* = *n* − 1, we obtain

$$
\omega(\mu) \ge \frac{\mu}{(n-2)!} \mu^{n-2} \omega^{(n-2)}(\mu),
\tag{21}
$$

eventually. Since *ω* is positive and increasing, we get that *x* ≥ (1 − *p*)*ω*, and so, (4) becomes

$$\frac{\mathbf{d}}{\mathbf{d}\mu} \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}\mu^{n-1}} \omega \right) + q \cdot \left[ F \circ (1 - \left[ p \circ \mathbf{g} \right]) \right] \cdot \left[ F \circ \omega \circ \mathbf{g} \right] \le 0,$$

which, with (21), yields

$$\frac{\mathbf{d}}{\mathbf{d}\mu} \left( \boldsymbol{a} \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}\mu^{n-1}} \boldsymbol{\omega} \right) + \boldsymbol{q} \cdot \left[ \boldsymbol{F} \circ (1 - \left[ \boldsymbol{p} \circ \mathbf{g} \right]) \right] \cdot \left[ \boldsymbol{F} \circ \frac{\mu \boldsymbol{g}^{n-2}}{(n-2)!} \right] \cdot \left[ \boldsymbol{F} \circ \boldsymbol{\omega}^{(n-2)} \circ \mathbf{g} \right] \leq 0.$$

Integrating this inequality from *u*<sup>1</sup> to *u*, we find

$$\begin{split} a(u)\omega^{(n-1)}(u) &\leq \ -\int\_{\mathfrak{u}\_{1}}^{u} q(\upsilon)F(1-p(\mathcal{g}(\upsilon)))F\left(\frac{\mu\mathcal{g}^{n-2}(\upsilon)}{(n-2)!}\right)F\left(\omega^{(n-2)}(\mathcal{g}(\upsilon))\right) \mathrm{d}\upsilon \\ &\leq \ -F(\omega^{(n-2)}(\mathcal{g}(\upsilon)))\int\_{\mathfrak{u}\_{1}}^{u} q(\upsilon)F(1-p(\mathcal{g}(\upsilon)))F\left(\frac{\mu\mathcal{g}^{n-2}(\upsilon)}{(n-2)!}\right) \mathrm{d}\upsilon. \end{split}$$

If we set *ψ* = *ω*(*n*−2) > 0, then we have that *ψ* is a positive solution of

$$(\psi'(u) + F(\psi(g(u))) \frac{1}{a(u)} \int\_{u\_1}^{u} q(\upsilon) F(1 - p(g(\upsilon))) F\left(\frac{\mu g^{n-2}(\upsilon)}{(n-2)!}\right) d\upsilon \le 0.$$

From Theorem 1 in [34], Equation (20) also has a positive solution, which is a contradiction. The proof is complete.

**Corollary 1.** *Assume that* lim*x*→<sup>0</sup> *<sup>x</sup> <sup>F</sup>*(*x*) = *K* < ∞*, F*(*x*)/*x* ≥ 1 *for* |*x*| ∈ (0, 1]*, and let (8) and (13) hold. The fulfillment of the following condition ensures the oscillation of Equation (4):*

$$\liminf\_{u \to \infty} \int\_{\mathcal{S}(u)}^u \frac{1}{a(s)} \left( \int\_{u\_1}^s q(\upsilon) F(1 - p(\mathcal{g}(\upsilon))) F\left(\frac{s \mathcal{g}^{n-2}(\upsilon)}{(n-2)!}\right) d\upsilon \right) ds > \frac{1}{\mathsf{e}}.\tag{22}$$

**Proof.** From Theorem 2.1.1 in [1], criterion (22) ensures the oscillation of Equation (20).
