I am readig Pugh's Analysis book:
Definition
Let $f:U \to \mathbb{R}^m$ be given where $U$ is an open subset of $\mathbb{R}^n$. The function $f$ is differentiable a $p \in U$ with derivative $(Df)_p = T$ if $T:\mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) \implies \lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$.
Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$\implies$" there rather than a "$\wedge$". Isn't it more natural to say
"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $\lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$"?
I'm having trouble seeing what the impact of changing these would be.
The author means
$$\forall R \left((\forall v ~ f(p+v) = f(p) + T(v) + R(v)) \implies \lim_{|v| \to 0} \frac{R(v)}{|v|} = 0\right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ \bigg(R = v \mapsto f(p + v) - f(p) - T(v)\bigg) \land \bigg(\lim_{|v| \to 0} \frac{R(v)}{|v|} = 0\bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v \mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$\begin{cases} \text{define } R \text{ as } v \mapsto f(p + v) - f(p) - T(v) \\ \lim_{|v| \to 0} \frac{R(v)}{|v|} = 0 \end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.