Given a true implication and false converse, what is the truth value of P↔¬Q? Here is the question:
Again, suppose the statement “if the square is blue, then the triangle is green” is true. This time however, assume the converse is false. Classify each statement below as true or false (if possible).
b. The square is blue if and only if the triangle is not green.

The correct answer is true, but I cannot see how that is the case. This is my thought process:
The statement is a biconditional and can be split into two parts.

*

*The square is blue if the triangle is not green (If the triangle is not green, then the square is blue).

*The square is blue only if the triangle is not green (If the square is blue, then the triangle is not green).

These two statements go against the original implication in the question. How can the square be blue, and at the same time be green and not green? My conclusion is that the statement is false. Is there a flaw in my deduction?
 A: You're running into the paradox of Material Implication.  You would indeed think that the statements:

"if the square is blue, then the triangle is green"

and:

"if the square is blue, then the triangle is not green"

cannot both be true, but remember that the logical (material) conditional has the rather unintuitive feature that as soon as the antecedent is false, the whole conditional is automatically true. So, if the square is not blue, then both statements are true. So these statements are in fact not contrary to each other.
Indeed, if we assume the two statement to be true, then clearly the square cannot be blue, because otherwise the triangle would be both green as well as not green. So, we know the square is not blue.
Even more straightforwardly: we are told the converse of:

"if the square is blue, then the triangle is green"

is false. The converse is:

"if the triangle is green, then the square is blue"

and for that to be false the triangle has to be green, and the square has to not be blue (again, it's a little counterintuitive, but material conditional can only be false when the antecedent is true and the consequent is false). With that, you immediately can figure out the truth-value of:

"The square is blue if and only if the triangle is green"

A: This truth table approach may be helpful:

Note the third row tells the story when $P\implies Q$ is true and $Q \implies P$ is false, where the fourth column from the left is the "logical and" of the statements, which is only true for row 3.
Comparing to the biconditional on the third row (depicted in the last four columns, with the result in the third column from the right under the $\iff$ symbol), one sees that it must be true as well.
I hope this helps.
A: You can verify this using the truth table. Note that "If P then Q" is equivalent to "$\neg P \lor Q$". Using B and G to represent "The square is blue" and "The triangle is green", respectively, you will conclude that "B" is False and "G" is True for the context of your question.
\begin{array}{cccccc}
 B & G & \neg B & \neg G& \neg B \lor G & \neg G \lor B \\
\hline
F&F&T&T&T&T\\
F&T&T&F&T&F\\
T&F&F&T&F&T\\
T&T&F&F&T&T
\end{array}
Now, "The square is blue if and only if the triangle is not green" is equivalent to "$(G\lor B)\land (\neg B\lor \neg G)$", which is true from the truth table.
A: The simplest thing to note is that if the converse ("If the triangle is green then the square is blue") is false the only way $P\implies Q$ can be false is if $P$ (the triangle is green) is true and $Q$ (the square is blue) is false.
So the triangle is green and the square is not blue.  That's the only option.
That means "if the square is blue then the triangle is green" is true (because $F\implies T$ is true).
.....
This also means "if the triangle is not green then the square is blue" is true (because $F\implies F$ is true).
And this also means "if the square is blue then the triangle is not green" is true (because $F\implies F$ is true).
So "the triangle is not green if and only if the square is blue" is true.
Now you divided the biconditional into two parts:

*

*The square is blue if the triangle is not green (If the triangle is not green, then the square is blue).

And as the triangle is green this is fine.  We can't conclude anything about the square.


*The square is blue only if the triangle is not green (If the square is blue, then the triangle is not green).

And as the triangle is green this means the square can't be blue. So we must conclude the square is not blue.
You ask  "How can the square be blue, and at the same time (the triangle) be green and not green?"
Well, it can't.
So the square is not blue.
A: Since the converse being false entails that the original statement is true, it is sufficient to just consider that the converse is false, i.e., $$\text{The triangle is green and the square isn't blue.}$$
As such, the given biconditional $$\text{The square is blue if and only if the triangle is not green.}$$ is vacuously true in both directions, so is true.
