Is the claim $(f\in P_1\cap P_2)\Rightarrow[(f\in P_3)\Leftrightarrow(f\in P_4)]$ identical to $(f\in P_1\cap P_2\cap P_3)\Leftrightarrow(f\in P_4)$?

Question

Let $A$ and $B$ be arbitrary finite sets, and let $f:A\to B$ be a function with domain $A$ and codomain $B$.

I am reading a paper whose main result reads like this:

1. Let the function $f:A\to B$ satisfy Properties 1 and 2. Then, the function $f:A\to B$ satisfies Property 3 if and only if it satisfies Property 4.

I am wondering whether the above statement is equivalent to the following one:

2. The function $f:A\to B$ satisfies Properties 1, 2 and 3 if and only if it satisfies Property 4.

Let me state the problem symbolically. Let $P_1$, $P_2$, $P_3$ and $P_4$ be arbitrary sets, where $f\in P_i$ if and only if $f$ satisfies Property $i$, where $i\in\{1,2,3,4\}$. I will now symbolically re-write statements 1 and 2:

1. $(f\in P_1\cap P_2)\Rightarrow[(f\in P_3)\Leftrightarrow(f\in P_4)]$.

2. $(f\in P_1\cap P_2\cap P_3)\Leftrightarrow(f\in P_4)$.

I have the suspicion that statements (1) and (2) are not equivalent, but I am failing to see how / where they differ.

Could you please help me figure out whether statements (1) and (2) are equivalent? If they are not, could you please help me see where / how they differ?

Answer 1

No, they're not equivalent. A function might satisfy both Properties 3 and 4 but not Properties 1 and 2; the first statement is vacuously true in that case, but the second statement is false in that case.

Restated in terms of sets, the first statement says that $P_1\cap P_2\cap P_3 = P_1\cap P_2\cap P_4$, while the second statement says that $P_1\cap P_2\cap P_3 = P_4$. These are not the same statement (unless $P_4 \subseteq P_1\cap P_2$). Correspondingly, the two original statements are not the same statement (unless Property 4 implies both Properties 1 and 2).