When "If $A$ is true then $B$ is true", is it valid to assert that "If $B$ is false, $A$ must also be false"? If it is given that: 

"People that ride buses, also ride planes"

then is the statement 

"people that don't ride planes, also don't ride buses"

necessarily true?
I don't think so, but the explanation to a problem in textbook I'm using uses that as logical proof to the answer provided.
It is based on the rule (according to this textbook) that, if this is true : ( if $A$ is true, then $B$ is also true), then it follows that if $B$ is false, then $A$ must also be false.

What you are asking is does:
A ⟹ B (A implies B)

mean
Not B ⟹ Not A (The converse of B implies the converse of A)?

The answer is yes.
In fact this is the basis of what is known as a Proof By Contradition in Maths.
The way I explain it to my A-Level students (18 year old Mathematicians in the UK) that meet this for the first time is with the following.
Let Statement A = "It is 12th of August 2018"

and
Let Statement B = "It is a Sunday"

Here
A ⟹ B

However, if it is not a Sunday (ie Not B, or the converse of B) is true, then we do not know what date it is with an absolute certainty, but we can say with absolute certainty it is NOT the 12th of August 2018, (ie we can say it is Not A, or we can say it is definitely the converse of statement A). It might be the 11th, it might be the 10th, but it definitely is not the 12th.
The famous one in Maths is the proof of the √2 cannot be expressed exactly as a fraction, ie √2 is irrational. This is done by a Proof by Contradition.
For contradition assume:
 A = "√2 is rational"

If that is true, then
 B = "√2 = a/b, where a and b are whole numbers, and 'a' and 'b' have no factors"

So we assume B is true and with a bit of basic maths we find if √2 = a/b, then both a and b have a factor of 2. So we have establish NOT B.
But from what we have said above:
If A ⟹ B, then Not B ⟹ Not A

So we have established that Not A is true, ie √2 is irrational, ie cannot be expressed as a fraction.
