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.