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Welcome to the forum. Problems about mathematical logic should be posted to Discrete Math subforum. In fact, see this post.

Oh, i've only just registered. Apparently, i've come late for the party!

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What is the scope of the existential quantifier in iii(a): just the premise $\displaystyle \neg(y\cdot x')=y$ or the whole implication. Please refer to syntactic conventions about omitting parentheses after the description of the syntax of first-order logic in your textbook. My guess is that the scope is the premise only, i.e., ∃ binds stronger than ->. Then the statement is true.

I'm afraid I only know as much (or less) as you do.

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Indeed, $\displaystyle \exists y\,y\cdot (x+1)\ne y$ is equivalent to x + 1 ≠ 1, i.e., x ≠ 0.

But only if $\displaystyle y \neq 0$?

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In fact, if the scope of ∃ extends until the end of the implication, the formula is still true. Indeed, for x = 0 we can take any y, e.g., y = 1. Then y(x + 1) ≠ y is false, so the whole implication is true. If x ≠ 0, then again we can take any y, and the implication is true in virtue of the truth of the conclusion.

I'll look at iii(b) and (c) later.

Ohhhh, I see! I had completely forgotten about that line in the implication truth table. I'll have another look at it tomorrow and post my new thoughts and answers.