$\forall b \space (b>1\implies b>5)$
is false since $b$ can be $2$ and $2 < 5$
It is correct?
Now,
if $b \in (1, \infty) \space$ $\implies$ $\space b > 5$
Is the second statement true or false or neither?
$\forall b \space (b>1\implies b>5)$
is false since $b$ can be $2$ and $2 < 5$
It is correct?
Now,
if $b \in (1, \infty) \space$ $\implies$ $\space b > 5$
Is the second statement true or false or neither?
Let $P_1$ and $P_2$ be two properties.
A statement of the form :
$$\forall (b) \Big[ P_1 (b) \implies P_2(b)\Big]$$
is equivalent to : there is no $b$ such that $b$ has the property $P_1$ and $b$ does not have the property $P_2$.
Note.- The literal meaning of an expression of this form is : for all $b$, if $b$ has property $P_1$, then, $b$ ( also) has property $P_2$.
Do you think there is no $b$ that
has the property of belonging to $(1, \infty)$
but that does not have the property of being strictly greater than $5$?
In case you can exhibit any number that has property $1$ but does not have property $2$, you've proved that the sentence is false.