Propositional Logic : Why is ((p $\Rightarrow$ r) $\land$ (q $\Rightarrow$ r)) $\equiv$ ((p $\lor$ q) $\Rightarrow$ r) I was working my way through some Propositional Logic and had the following doubt :

Why is this true :
((p $\Rightarrow$ r) $\land$ (q $\Rightarrow$ r)) $\equiv$ ((p $\lor$ q) $\Rightarrow$ r)

Please provide an intuitive explanation and not one that uses a truth table or logic identities to simplify the expression . I have already done both of them :)
 A: From $\;((p\implies r) \wedge (q\implies r))\;$, we know if $p$ is true then $r$ is true, and if $q$ is true then $r$ is true.  
This means that if either $p$ or $q$ are true then $r$ is true.  (If one is false and the other true, we still know $r$ must be true; only if both are false do we not know what $r$ is.)
Which is $\;(p\vee q) \implies r\;$ so they are equivalent statements.
$$\therefore ((p\implies r) \wedge (q\implies r)) \equiv ((p\vee q) \implies r)$$

Alternatively, since $\;x\implies y\;$ is equivalent to $\;\neg x \impliedby \neg y\;$ then we can substitute a contraposition for each implication.
Thus $\;((p\implies r) \wedge (q\implies r))\;$ is equivalent to $\;((\neg p\impliedby \neg r) \wedge (\neg q\impliedby \neg r))\;$, and hence we know if $r$ is false then $p$ is false, and if $r$ is false then $q$ is false.  
To simplify, this means that if $r$ is false then both $p$ and $q$ are false.
Which is $\;(\neg p\wedge \neg q) \impliedby \neg r\;$, and this is equivalent to: $\;(p\vee q) \implies r\;$.
$$\because ((\neg p\impliedby \neg r) \wedge (\neg q\impliedby \neg r)) \equiv ((\neg p\wedge \neg q) \impliedby \neg r)$$
$$\therefore ((p\implies r) \wedge (q\implies r)) \equiv ((p\vee q) \implies r)$$

Gregor de Cillia put this rather more succinctly in the comments. 

If every penny is a remedy and every quarter is a remedy, then everything that is a penny or a quarter is a remedy. The inverse can be stated similary –  Gregor de Cillia 

A: Here is a no nonsense approach that should be quite easy to follow:
\begin{align}
[(p\to r)\land(q\to r)] &\equiv [(\neg p\lor r)\land (\neg q\lor r)]\tag{$p\to q \equiv \neg p \lor q$}\\[0.5em]
                        &\equiv [(\neg p\land \neg q)\lor r] \tag{distributivity}\\[0.5em]
                        &\equiv [(\neg(\neg p\land \neg q)\to r]\tag{$p\to q \equiv \neg p \lor q$}\\[0.5em]
                        &\equiv (p\lor q)\to r\tag{DeMorgan}
\end{align}
Thus, we have that 
$$
[(p\to r)\land(q\to r)]\equiv (p\lor q)\to r,
$$
as desired.
A: As noted by @GregordeCillia in the comments , the intuitive explanation would be something on these lines :

  
*
  
*If every penny is a remedy and every quarter is a remedy $\equiv$ ((p $\Rightarrow$ r) $\land$ (q $\Rightarrow$ r))
  
*then everything that is a penny or a quarter is a remedy $\equiv$ ((p $\lor$ q) $\Rightarrow$ r)
  
*The inverse can be stated similary 

