Which rule of logic is used on $\delta$ in this proof of the squeeze theorem In the Squeeze Theorem delta epsilon proof, we get to a step where
$$
\begin{align*}
& \text{Let } \delta = \min(\delta_1,\delta_2,\delta_3). \\
& \text{Then by }(1), (3),\text{ and }(2) \\
\end{align*}
$$
I was wondering which logical law is being employed on this step to let you replace like:
$$
\forall x[ |x - c| < \delta_1 \rightarrow \dots], \delta_1 \geq \delta \vdash \forall x[ |x -c| < \delta \rightarrow \dots]
$$
Which logical or set laws are being employed here?  One thing I could observe is that you're reducing a set to its subset when you replace $\delta_1$ with $ \delta $.  But I'd like to know formally what laws are being applied to do the delta replacement here.. So I could for example label it on the right side of the proof when I apply the inference rule.
 A: The proof of that assertion can be made using the following reasoning.
First, suppose the following two facts:
$$\forall x\left[|x-c|<\delta_1\implies\dots\right]\tag{1}$$
$$\delta_1\geq\delta\tag{2}$$
Fact $(2)$ follows from the definition of the minimum as you are aware.
Let $x\in\mathbb{R}$.
$$|x-c|<\delta\implies|x-c|<\delta_1\tag{3}$$ as $\delta\leq\delta_1$. This follows from the transitivity axioms of an ordered field.
By universal instantiation applied to $(1)$,
$$\left[|x-c|<\delta_1\implies\dots\right]$$
But note that
$$|x-c|<\delta\implies|x-c|<\delta_1\implies\ldots$$
By the transitivity of implication, we have that:
$$|x-c|<\delta\implies\ldots$$
The next inference rule we will be using is called "universal generalisation", which states that:
$$[z \text{ arbitrary and } P(z)] \vdash\forall a\;P(a)$$
However, $x$ was arbitrary, therefore the following holds by universal generalisation:
$$\forall x \left[|x-c|<\delta\implies\dots \right]$$
Rounding out the answer in the notation that you're using, we have:
$$\forall x\left[|x-c|<\delta_1\implies\dots\right]\vdash \forall x \left[|x-c|<\delta\implies\dots \right]$$
A: Much thanks to @Kcronix for laying out the logical steps. Here is my attempt.
Squeeze Theorem: Suppose that $$ f(x) \leq g(x) \leq h(x)$$ for all $x$ in some interval around $c$, with the possible exclusion of $c$ itself. Also suppose that $$ \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x). $$ Then $$\lim_{x\to c} g(x) = L.$$
Proof: We want to prove $\lim\limits_{x\to c} g(x) = L$ given the premises above it.  Let $\epsilon$ be any real number more than $0$, and let $x$ be any real number.  By the definition of a limit,
$$ \lim_{x\to c}f(x) = L$$
means there exists a a $\delta_f$ such that,
$$ 
\begin{align}
|x-c| < \delta_f &\rightarrow |f(x) - L| < \epsilon \tag{1} \\
|x-c| < \delta_f &\rightarrow -\epsilon < f(x) - L < \epsilon.
\end{align}
$$
And
$$ \lim_{x\to c}h(x) = L$$
means there exists a a $\delta_h$ such that,
$$
\begin{align}
|x-c| < \delta_h &\rightarrow |h(x) - L| < \epsilon \tag{2} \\
|x-c| < \delta_h &\rightarrow -\epsilon < h(x) - L < \epsilon.
\end{align}
$$
The open interval around $c$, with the possible exclusion of $c$ itself, can be described with some $\delta_g > 0$, such that
$$ 
\begin{align}
|x-c| < \delta_g &\rightarrow f(x) \leq g(x) \leq h(x) \tag{3} \\
|x-c| < \delta_g &\rightarrow f(x) - L \leq g(x) - L \leq h(x) - L. \\
\end{align}
$$
Let $\delta$ be the minimum of $\delta_f$, $\delta_h$, and $\delta_g$.  Then by the transitivity of inequality, we can substitute $\delta$ in for the previous deltas in (1), (2), and (3).  So with our $\delta > 0$ we have:
$$
\begin{align}
|x-c| < \delta_g \rightarrow& -\epsilon < f(x) - L < \epsilon, \\
& -\epsilon < h(x) - L < \epsilon, \\
& f(x) - L \leq g(x) - L \leq h(x) - L \\
|x-c| < \delta_g \rightarrow& -\epsilon < f(x) - L < \epsilon
\end{align}
$$
So by the definition of a limit:
$$ \lim_{x\to c} g(x) = L. $$
A: I suppose you'd say that if you know $\forall x \in A, P(x)$ for $P$ some property, then whenever $B \subset A$ then you automatically know $\forall x \in B, P(x)$, too. So in a "forall" passing to quantifying over a subset is always OK.
But let me be unhelpful and suggest that logical manipulations going in this proof are not so important (depending on your purposes for going through it) compared to the big picture. Often $\varepsilon$-$\delta$ arguments are a bit hard or at least tedious to read, and equally as often there are ways to avoid them: here's a one way to hide the "$\varepsilon$"s in this case using a little bit of technology.
First you define the $\limsup_{x \to c} f(x)$ (called the limit superior) and $\liminf_{x \to c} f(x)$ (called the limit inferior), which are slight variations on the ordinary limit which you can look up in probably any book on real analysis. (The intuition is that the limit superior is the largest value which it looks like $f(x)$ is attaining as $x \to c$, and likewise the limit inferior is the smallest such value, but I'd suggest you read the definition and then think about this.) The upshot is that:

*

*Unlike ordinary limits the $\limsup$ and $\liminf$ always exist (if we allow $\pm \infty$), we always have $\limsup_{x \to c} f(x) \geq \liminf_{x \to c} f(x)$, and the limit $\lim_{x \to c} f(x)$ exists exactly when $\limsup_{x \to c} f(x)$ and $\liminf_{x \to c} f(x)$ are equal.


*Moreover it is super easy to prove from the definition of $\limsup$ and $\liminf$ (exercise!) that whenever $f(x) \leq g(x)$ then $\limsup_{x \to c} f(x) \leq \limsup_{x \to c} g(x)$ and likewise for the limit inferior.
I'll leave it to you to then prove the squeeze theorem by combining (1.) and (2.)---we're basically done already!
