Is the biconditional possible for this statement (instead of just a conditional)?

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In one of the assignment of the "Introduction to Mathematical Thinking" course by professor Keith Devlin on Coursera, this statement was shown to be true:

$\forall x \forall y \, [(x \leq y) \land (y \leq x) \Rightarrow (x = y)]$

However, it seems it can go in the reverse direction as well:

$\forall x \forall y \, [(x = y) \Rightarrow (x \leq y) \land (y \leq x)]$

It seems quite natural to me that this is necessarily true, but could someone help confirm this?

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Yes, the statement $\forall x \forall y \, (x = y \to (x \leq y \land y \leq x))$ also holds. Indeed, $x \leq y$ is equivalent to say that either $x < y$ or $x = y$. So, in particular, $x = y$ implies that $x \leq y$ (and similarly that $y \leq x$).