Universal quantifier question: $(\forall x)[x < 0 \Rightarrow x^2 > 0]$ Universal quantifier question: $(\forall x)[x < 0 \Rightarrow  x^2 > 0]$. 
Given the above expression, For all of $x$ [ if $x$ is less than zero, then $x^2$ is greater than zero].
Is that a true statement?
The confusion: $p \Rightarrow q$, is true in all cases except when $p$ is true and $q$ is false. In the above problem, if $x < 0$  is true then $x^2 > 0$ will be true. But, when, $x = 0$, then, the premise $x < 0$ is false. But, according to definition, if premise is false, the expression is still considered true. Doesn't it contradict?
 A: 
The confusion: p → q, is true in all cases except when p is true and q is false. In the above problem, if x < 0 is true then x² > 0 will be true. But, when, x = 0, then, the premise x < 0 is false. But, according to definition, if premise is false, the expression is still considered true. Doesn't it contradict?

Where the antecedent of an implication is false, it is the statement that is held to be true; not necessarily its consequent.
That is, an implication statement is true if: either the antecedent and the consequent are both true, or the antecedent is false (whatever the consequent may be).   Which is the same as saying that an implication statement is considered false only if: the antecedent is true while the consequent is false.
Hence, the statement $\forall x \;(x<0 \,\to\, x^2> 0)$ will be contradicted only by an example, $c$, such that $c<0$ and $c^2\leq 0$.   The example of $0$ does not contradict the statement.



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*"All ducks fly south for the winter" is not contradicted by non-migrating emus.


$$\forall \text{bird}\;(\text{bird}\in\text{Ducks}\;\to\;\text{bird}\in\text{Southward winter migrators}) \\ \text{Emu}\not\in\text{Ducks}\;\wedge\;\text{Emu}\not\in\text{Southward winter migrators} \\ \ldots$$


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*"All swans are white" was contradicted by finding black swans.


$$\forall \text{bird}\;(\text{bird}\in\text{Swans}\;\to\;\text{bird}\in\text{White feathered}) \\\text{Black Swan}\in\text{Swans}\;\wedge\;\text{Black Swan}\not\in\text{White feathered}\\\text{Contradiction!}$$
