I am currently self-studying propositional logic from the 4th Edition of Mathematical Proofs: A Transition to Advanced Mathematics. Having finished the chapters on negation, implication and biconditionals, I find myself a bit confused, and I believe the best way for me to express that confusion would be from the following example:
Let the following be open sentences, across the domain of natural numbers:
P(x): x is prime.
Q(x): x is equal to 2.
R(x): x is odd.
Now, I’m sure it would be correct to say that:
P(x) and ~Q(x), implies R(x).
However, I’m wondering whether it would be correct to say that:
P(x) and ~R(x), implies Q(x).
Can this even be considered an implication, considering that there is no situation in which P(x) and ~R(x) are false while Q(x) is true? It seems like more of a biconditional, since for one to be true, the other must be true as well, and if one is false, the other one must be false too- however, there’s also no situation in which P(x) and ~R(X) are true while Q(X) is false, so wouldn’t that make the implication valid as well?
To summarize my question here, I’m just wondering- can a biconditional be considered implication? If one situation depends on another to be true, and vice versa, does that still mean that one implies the other? Thanks in advance.
Let $P$ and $Q$ be complex formulas. Then, by definition, a biconditional $P \leftrightarrow Q$ is logically equivalent to $(P \to Q) \wedge (Q \to P)$. In other words, a biconditional is logically equivalent to the conjunction of a conditional $P \to Q$ and its converse $Q \to P$.
Furthermore, if $(P \to Q) \wedge (Q \to P)$ holds, then we can derive either $P \to Q$ or $Q \to P$ on a separate line of proof by conjunction elimination, also known as simplification. So, given a biconditional like the one above, it is perfectly correct to conclude that $P$ implies $Q$. It is also perfectly correct to conclude that $Q$ implies $P$. That is because the biconditional operator $\leftrightarrow$ implies that both $P$ and $Q$ imply each other.
So as for your example, if we have...
$P(x)$: $x$ is prime.
$Q(x)$: $x$ is equal to $2$.
$R(x)$: $x$ is odd.
then you are correct in saying that, for every natural number $x$...
$( P(x) \wedge \neg R(x) ) \to Q(x)$
You are also correct in saying that, for every natural number $x$...
$Q(x) \to ( P(x) \wedge \neg R(x) )$
So, by definition of biconditional, it is also correct to state that, for every natural number $x$...
$( P(x) \wedge \neg R(x) ) \leftrightarrow Q(x)$
If the first two expressions hold, then so does the third. And if the third expression holds, then so do the first two.