why the statement $((p ∨ r) → q) ↔ ((p → q) ∨ (r → q))$ is not always true I can't intuitively  see why the statement $((p ∨ r) → q) ↔ ((p → q) ∨ (r → q))$ is not always true. When I fill out a truth table I see that $((p ∨ r) → q)$ is not always equal to $((p → q) ∨ (r → q)),$ but it still really "feels" like these two statements say the same thing. Can you help me see why, in an intuive way, rather than a proof/truth table?
 A: Since you are looking for intuition here is something that might help, also related to the distribution property not holding for the implication, as stated above. Consider the statements to be:
$p:$ "there is precipitation".
$r:$ "the temperature is below $0$ degrees Celsius".
$q:$ "it will snow".
The combination of $p$ and $r$ implies $q$ (meaning that $p\wedge r \to q$), in particular if $p$ is true but $r$ and $q$ are false, then $p ∨ r \to q$ has to be false: since there is precipitation yet no snow. This is clearly not the case for  $p \to q ∨ r \to q$, which is true since $r \to q$ holds true, we see no snow and no temperature below $0$ (false implies false is true).
A: I think the reason why it feels like these are equivalent is this:
You are looking at the LHS, and say: "OK, so I know that $r$ is true if $p \lor q$ is true. OK, but that then means that if $p$ by itself is true, we have $r$ as well. Alternatively, if $q$ by itself is true, we get $r$. So, we're going to get $r$ when ('if'!) $p$ is true, or ('alternatively') when ('if'!) $q$ is true."
OK, and now notice how close that last statement is to the RHS ...
Of course, what you should say is that $r$ is true when $p$ is true, and that $r$ is true when $q$ is true. But, in English we kind of contract that. We say: "whether this or whether that, bla bla".
Indeed, consider the following statement:
'Apples and oranges are fruits'
Now, if you use $A$ for 'you have an apple in your hand', $O$ for 'you have an orange in your hand', and $F$ for 'you have a fruit in your hand', you might be inclined to translate this as:
$(A \land O) \to F$
But that is of course wrong!  Now it is saying that if you have an apple and an orange in your hand, then you have a fruit.  What we should be saying instead is $(A \lor O) \to F$. OK, so where does the 'and' in 'apples and oranges are fruit' come from then?  It is because we just contracted the statement $(A \to F) \land (O \to F)$ that is equivalent to this.
A second thing that may be at play here is that the LHS does logically imply the RHS, and we are naturally looking and thinking about equalities going from left to right, more so than right to left, we may have the illusion that we are dealing with an equivalence, rather than just a one-way implication.
Indeed, if you start with the RHS, you are far less likely to come up with the LHS. That is, starting with the RHS, I think you realize that just because $r$ is true just because $p$ is true, does not mean that we suddenly get $r$ just because either $p$ or $q$ is true: we realize that all we know is that $r$ is true when $p$ is true, so why would it be true in some different case?  And, likewise, if we just have that $r$ is true whenever $q$ is true, we don;t suddenly have that $r$ is true when either $p$ or $q$ is true.  So, starting from the RHS we (correctly!) realize that we cannot end up at the lHS.
So, in the end I think it is partly of how we're used to using English, the kinds of contractions we make in English, and how that guides our thinking, together with the fact that we look at statements like this with a left-to-right reading 'bias'.
A: Hınt: Remember that implication does not have distribution property. $$((p \lor r)\rightarrow q) \equiv (p \rightarrow q) \land (r \rightarrow q)$$
For intuitive approach for OP: We know that the "implication" can be seen as an "agreement".For example: Lets say:
$p:$ we meet at park ,
$q:$ we will eat ice-cream ,
so $(p \rightarrow q):$ if we meet at park ,we will eat ice cream.
We know that this "agreement" is only false in such case : when we meet at park but we do not eat ice cream.
Then , lets think it for $((p \lor r)\rightarrow q)$ , lets say $r:$ we meet at school.
So , $((p \lor r)\rightarrow q):$ If we meet at park or school,we will eat ice cream. It is apparent that this can be translated such a way that : "if we meet at park , we will eat ice cream ,and if we meet at school,we will eat ice cream"
A: $$\Big((p ∨ r) → q\Big) ↔ \Big((p → q) ∨ (r → q)\Big)\tag1$$

*

*Statement $(1)$ is guaranteed true in any world where all its three conditionals are true, in other words, where the consequent $q$ is true.


*On the other hand, consider the interpretation $\quad p :=$ a circle is
angular, $\quad q :=$ a circle has sides, $\quad r :=$
a circle is round.
Statement $(1)$'s RHS $$(\color{red}p → q) ∨ (r → q)$$ is
immediately true by virtue of $\color{red}p$ being false.
But statement $(1)$'s LHS $$(p ∨ r) → q$$ is false, since its
premise is true but its conclusion false.
A: Intuition is very subjective, so I'll base my one on formality :)
There is only 2 cases where left side differs from right and, as expected, they are symmetric: $(p,q,r) = (1,0,0)$ and $(p,q,r) = (0,0,1)$. Possibly, "intuitively",  we can say:
Taking the preconditions $p,q$ together, in one "or", lets mess up the sentence i.e.  makes it false, while taking them  separately does not give the opportunity to spoil the whole sentence i.e. it is always true.
Don't Put All your Eggs in One Basket.
A: An example where the two sides differ could be:
Q: The house is NOT for sale.
So, if Q is true, there is no possible trade.
If Q is false (i.e. the house IS for sale), we have two parties, P, the seller, and R, the buyer.
If P wishes to sell and R wants to buy, the trade is over.
If P doesn't want to sell and R doesn't want to buy, the trade is over.
On the LHS, both other possibilities close the trade, whereas on the RHS both other possibilities leave the trade still open.
