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.

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: $$\forall x\left[|x-c|<\delta_1\implies\dots\right]\vdash \forall x \left[|x-c|<\delta\implies\dots \right]$$