Proof expression $((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D$ by rules of interference

Proof expression $$((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D$$

Attempt to proof

$$1. B \text{ (premise) }$$ $$2. B \rightarrow D (\text{premise})$$ $$3. A \rightarrow D (\text{premise})$$ $$4. D (\text{modus ponens})$$ $$5.(\neg A \vee D) Add$$

Is this correct?

$X \rtimes Y \simeq X \vee (X \wedge Y)$ for $X$ a Co-H-Space

I have asked the below question on MathSE (with a 200 point bounty) but have yet to recieve an answer there, and so am trying here. I am happy to remove it if it is nevertheless decided that this question is not appropriate here.

Let $$X$$ and $$Y$$ be pointed CW complexes, with $$X$$ a Co-H-Space with co-multiplication $$\mu$$.

I believe that the following identity holds.

$$X \rtimes Y \simeq X \vee (X \wedge Y),$$

where $$\rtimes$$ denotes the half-smash: $$X \rtimes Y := X \times Y/(* \times Y)$$.

I am looking for a proof, or a reference for one.

I am aware that this identity holds when $$X$$ is a suspension. A technique for proving it in that case is to prove that $$\Sigma A \rtimes Y \simeq \Sigma(A \rtimes Y)$$, by taking the homotopy pushout of the diagram $$* \leftarrow X \rightarrow *$$, taking the product with $$Y$$ everywhere and quotienting by $$Y$$ everywhere, leaving us with a diagram that is still a homotopy pushout diagram.

It doesn’t seem that this argument can be applied here, though.

The co-multiplication $$\mu$$ on $$X$$ induces a co-multiplication $$\bar{\mu}$$ on $$X \rtimes Y$$ (and also on $$X \wedge Y$$). This gives us an obvious map:

$$\phi:=(p_1 \vee q) \circ \bar{\mu}:X \rtimes Y \rightarrow (X \rtimes Y) \vee (X \rtimes Y) \rightarrow X \vee (X \wedge Y)$$

where $$p_1$$ and $$q$$ are the projection $$X \rtimes Y \rightarrow X$$ and the quotient map $$X \rtimes Y \rightarrow X \wedge Y$$, respectively.

I cannot see how it could be deduced that this map is a homotopy equivalence, or if it is even the map I want.