# Help understanding $x=y\Rightarrow(x=z\Rightarrow y=z)$

I was reading a proof that opened with the integer axiom of $x=y\Rightarrow(x=z\Rightarrow y=z)$

What would be an accurate statement in English to express this? The "implies" within the first "implies" is kind of confusing to me. I believe the general idea is that if $x$ equals $y$, then if $x$ equals $z$, $y$ also equals $z$.

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If $x$ is equal to $y$, then anything equal to $x$ must also be equal to $y$.

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"If z is the same thing as x, and y is the same thing as x, then z is the same thing as y."

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...and this is essentially using the fact that $\;P \Rightarrow (Q \Rightarrow R)\;$ is equivalent to $\;(P \land Q) \Rightarrow R\;$. –  Marnix Klooster Aug 4 at 6:04

In general, you can use the word "that" to describe implication of implications:

$x=y$ implies that $x=z$ implies $y=z$.

"p implies q" implies $that$ if not q then not p.

$(p \implies q) \implies (\neg q \implies \neg p)$

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In my opinion this is most easily understood as a specific instance of $$(0) \;\;\; x = y \;\Rightarrow\; (P(x) \Rightarrow P(y))$$ where $\;P(\cdot)\;$ is any boolean expression of one variable. And in turn this is a weaker form of $$(1) \;\;\; x = y \;\Rightarrow\; (P(x) \equiv P(y))$$ which Dijkstra calls Leibniz' rule, and which the Wikipedia page on the topic calls "The indiscernibility of identicals".
Now $(1)$ in plain English: "If $\;x\;$ equals $\;y\;$, then whatever is true of $\;x\;$ is true of $\;y\;$, and vice versa." (If you leave out the "and vice versa" part, of course you get $(0)$.)