Within If and Only If Proofs, why can the proof for one direction be easier than the other? For $ P \iff Q$, my initial sentiment is that because P and Q are equivalent, the total of two proofs (one for each direction) should entail the equivalent level of "difficulty" or "exertion", as well as an analogous number of steps. Otherwise, they wouldn't be equivalent.
Yet, this is false and I can't perceive why? 
Moreover, a priori, how would one divine/previse:
$2.$ if one direction is truly easier than the other  ?   
$3.$ then which of the two is it, on the condition that $2$ is true?
 A: I don't think there is a reasonable general answer to this very genuine phenomenon: that's just the way the (mathematical) world is.
In a related area, most of modern cryptography (and thus of the modern economy: think banking) relies on the fact that it is very easy to multiply two huge prime numbers $p, q$ but extremely difficult, given just their product $n=p.q$, to retrieve the two factors $p,q$.
A: Well, as you know, Fermat’s Last Theorem has been proven. Therefore you don’t lose any truth if you modify the statement “$1 + 1 = x$ if and only if $x = 2$” to “$1 + 1 = x$ if and only if ($x = 2$ and, yeah, Fermat’s Last Theorem is true)”. This doesn’t have to happen explicitly. You can easily add a lot of implicit statements to one side. This may be one reason why directions of equivalences are not equally easy to prove.
For any equivalence whose directions are equally easy to prove, there a many modifications breaking the symmetry of difficulty. So if you assume that equivalences of the former kind exist, mathematicians are apparently more interested in their asymmetric modifications, and it is hard to see what the “pure equivalences” behind them are.
But I don’t think it’s helpful to think of equivalences as modifications of pure equivalences (whose direcetions are equally easy to prove) because there’s probably no sensible way of determining whether an equivalence is pure or not (maybe the only pure equivalence is “true if and only if true”?) – I was just addressing your feeling that it should be something like that.
A: I do not "see" any "logical" reason why - in general - the two "directions" of the equivalence poof must have different "levels of difficulty"...
(i) how is "measured" the "level of difficulty" ?
(ii) we do not need necessarily divide the equivalence proof into two subproofs: one for each "direction"; nothing prevents us - in general - to work with a "chain of equivalences" (there is a proof systems, named Equational Logic that is based exactly on chains of equivalences [but see George Tourlakis, Mathematical Logic (2008), for a rigorous treatment].
A: Because you are proving two things which happen to be converses. Given two theorems, often one is harder than the other. This still is true when they are converses because being the converse of another statement doesn't give you anything special.
A: The "difficulty" of a proof is a subjective concept - it is contingent on the way our brains work. Thus, even though $p \Leftrightarrow q$ it may be harder to prove one direction than the other because of the way humans think. 
That would be like saying in a certain equivalence proof, people really enjoy proving $p \implies q$ but people really hate proving $q \implies p$ and then arguing that since the two statements are equivalent we should either hate proving both directions or love proving both directions...
A: Obviously it is not necessary that two directions have the  same level of difficulty; Like when one direction is a special case of the other.For example when we want to prove that a homomorphism is 1-1 we only need to prove at one point:0
A: Let us analyze your question from an intuitive point-of-view. In the so called deductive systems, loosely speaking, you have a set of axioms, some rigorous and logical method of proofs and then you deduce theorem after theorems. Right?
But the subtle point in absolute rigor is that you deduce your results basing only upon previously proved theorems. Thus you get an 'ordered chain' of rigorously proved theorems. You may define the Degree of Complexity of theorems as a function natural numbers. For example, you may define the DC of each of the axioms of your mathematical system as $1$ and then you may define the DC of the theorem $T$ (denoting it by $\delta (T)$) as $\delta (T)$ $=$ $\upsilon (T)$ where $\upsilon (T)$ denotes the number of use of the axioms. So in this view, if for two propositions $P$ and $Q$ we have $\upsilon(P)$ $=$ $\upsilon (Q)$ then the two propositions will be said to be Equally Complex Theorems and if not then the greater $\upsilon(x)$ will denote a theorem of greater complexity. 
But as I have said earlier, this is not a rigorous approach to your question.It is also obvious from the representation because I have not proved that why there will be a functional correspondence between $\mathbb N$ and your mathematical system. However you may think of your problem in the way I have described above.
