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Look at this proof: Uniqueness Proof, Discrete Math Help

The first proof is invalid right? There is a mistake when proving uniqueness, I think.

They have proved that there exists $r$ s.t $ar+b=0$, and to prove uniqueness they assume there exists $s$ s.t $as+b=0$, BUT then they just assert $ar+b=as+b$

Here are the three strategies for proving uniqueness: enter image description here

And in none of them you can assume that $P(y) = P(z)$

Why is everyone in the comments saying that it's valid?

enoopreuse22
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    Note that $0=0$ (even if its obvious). Indeed $ar+b=0$ and $as+b=0$ thus $ar+b=as+b$. – Lelouch May 22 '23 at 19:48
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    (a) Why aren't you commenting on the other question, rather than asking a new question? (b) Your $P$ is intended to be a property that uniquely defines the value in question, so $P(x)$ means $a x + b = 0$ in the example in the other question. The proof in the other question does not assume $P(r) = P(s)$ (whatever that means), it assumes $P(r) \land P(s)$, i.e, $ar+b = 0$ and $as +b = 0$, implying that $ar+b = as+b$. This is in line with with all of your strategies for proving uniqueness. – Rob Arthan May 22 '23 at 20:00

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