is $(\phi^2 > 2) \Rightarrow (\phi > 1.4)$ true or false? First, I need to evaluate left and right hand sides of '$\Rightarrow$' to use definition of implication (its truth table).
And, I'm simply lost in a question: "how should I evaluate truth or falsity of both $(\phi^2 > 2)$ and $(\phi > 1.4)$ ?
Both of left hand side and right hand side of implication depend on $\phi$
And any reasoning I came up with looks like: there will be intervals of $\phi$ where this statement is true and intervals where this statement is false...
We seek when LHS is True:  $(\phi < - \sqrt 2) \cup (\phi > \sqrt 2)$. Then $\phi > 1.4$ is true for $\phi > \sqrt 2$ but false for $\phi < - \sqrt 2$.
So entire expression is True $\phi > \sqrt 2$ and false for $\phi < - \sqrt 2$
Now consider $ 1.4 < \phi <= \sqrt 2$. RHS is True and LHS is true.
So entire expression is true.
For $ -\sqrt 2 \le \phi \le 1.4$, RHS is false and LHS is false as well, so entire expression is true.
So the final answer is:
$(\phi^2 > 2) \Rightarrow (\phi > 1.4)$ is false for $\phi < - \sqrt 2$ and true for $\phi >= - \sqrt 2$
Did I get it right?
 A: Assuming that $\phi$ is an arbitrary real number, then the assertion is false, since, for instance, $(-2)^2>1.4$, but it is not true that $-2>1.4$.
A: As you said, you will have four open sub-intervals given by:
$$
I_1 = \left(-\infty, -\sqrt{2}\right)
$$
$$
I_2 = \left(-\sqrt{2}, \ 1.4\right)
$$
$$
I_3 = \left(1.4, \ \sqrt{2}\right)
$$
$$
I_4 = \left(\sqrt{2}, \ \infty\right)
$$
$$
I = I_1 \cup I_2 \cup I_3 \cup I_4 = \mathbb{R}
$$
Let's enumerate the statements $(1)$, $(2)$ and $(3)$
$$
\phi^2 > 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
$$
\phi > 1.4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
$$
$$
\left(1\right) \Rightarrow \left(2\right)  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
$$
The statement $(1)$ is true for $T_1$ and false for $F_1$
$$
\phi^2 > 2 \Rightarrow 
\begin{cases}
\phi < -\sqrt{2} \\
\phi > \sqrt{2}
\end{cases} \Rightarrow \begin{cases}
T_1 = I_1 \cup I_4 \\ F_1 = I_2 \cup I_3
\end{cases}
$$
The statement $(2)$ is true for $T_2$ and false for $F_2$
$$
\phi > 1.4 \Rightarrow \begin{cases}
T_2 = I_3 \cup I_4 \\ F_2 = I_1 \cup I_2
\end{cases}
$$
Then if we want to analyze what $(3)$ is, we want to know our objective.
Case 1: Know if $(3)$ is true for every $\phi$
$$
(1) \Rightarrow (2) \ \ \ \ \ \ \ \ \ \text{if} \ \ \ T_1 \subseteq T_2
$$
As we have
$$
\left(I_1 \cup I_4\right) \not\subseteq \left(I_3 \cup I_4\right)
$$
Then the $(1) \Rightarrow (2)$ is False.
Case 2: Search for which values of $\phi$ the statement $(3)$ is valid.
The statement $(3)$ is true for $T_3$ and false for $F_3$. Then
$$
T_3 = \left[T_1 \cap T_2\right] \cup \left[F_1 \right]
$$
$$
T_3 = \underbrace{\left[\left(I_1 \cup I_4\right) \cap \left(I_3 \cup I_4\right)\right]}_{I_4} \cup \left[I_2 \cup I_3\right]
$$
$$
\boxed{T_3 = I_2 \cup I_3 \cup I_4}
$$
$$
F_3 = I \setminus T_3 = I_1
$$
That means
$$
T_3 = \left(-\sqrt{2}, \ \infty\right)
$$
$$
F_3 = \left(-\infty, \ -\sqrt{2}\right)
$$
PS: I took out the exact values of $-\sqrt{2}$, $1.4$ and $\sqrt{2}$ just to make it easier to write. But for a more formal analysis you should include them.
EDIT: For the case
$$
(1) \Leftrightarrow (2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
$$
Then
$$
T_4 = \left[T_1 \cap T_2\right] \cup \left[F_1 \cap F_2\right]
$$
$$
T_4 = \left(-\sqrt{2}, \ 1.4\right) \cup \left(\sqrt{2}, \ \infty\right)
$$
$$
F_4 = \left(-\infty, \ -\sqrt{2}\right) \cup \left(1.4, \ \sqrt{2}\right)
$$
A: In the standard interpretation (where the symbols have their usual arithmetical meanings) and with domain of discourse $\mathbb R,$ $$\forall \phi\;\Big((\phi^2 > 2 \to \phi > 1.4)\iff \phi\in[-\sqrt2,\infty)\Big).\tag1$$ So, for $$\phi^2 > 2 \implies \phi > 1.4$$ to be universally true, we can change the interpretation by changing the domain of discourse, by restricting it from $\mathbb R.$ There are infinitely many ways of doing this, and from $(1),$ the most conservative restriction is $$\mathbb R{\setminus}(-\infty,-\sqrt2).$$

Addendum
Here's one way of obtaining $(1):$
\begin{align}&\phi^2 > 2 \to \phi > 1.4
\\\iff{}&\phi^2 \le 2 \quad\lor\quad \phi > 1.4
\\\iff{}& -\sqrt2\le\phi \le \sqrt2 \quad\lor\quad \phi > 1.4
\\\iff{}& -\sqrt2\le\phi
\end{align}
