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This question already has an answer here:

Suppose that f is a function from A to B, where A and B are finite sets |A| = |B|. Show that f is one-to-one if and only if it is onto.

Lets say we were to translate this statement using variables. Where:

a: f is a function from A to B, where A and B are finite sets |A| = |B|

b: f is one-to-one

c: f is onto

Now, I believe this statement translates to "a -> (b <-> c)" symbolically. Is this correct?

This would mean that we need to prove the two sub-statements "a -> (b -> c)" and "a -> (c -> b)" in order to prove the overall statement "a -> (b <-> c)".

To prove "a -> (b -> c)", I'm thinking of first showing that "b -> c" is true using a proof by contradiction. We would then get "a -> T", which is always true (trivial proof). Is this correct?

To prove "a -> (c -> b)", we would use the same procedure.

Is this what the top answerer here does?

Questions highlighted in bold.

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marked as duplicate by Jack's wasted life, John B, Namaste algebra-precalculus Oct 18 '16 at 10:54

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