Is the truth table of $(p↔q)↔(r↔s)$ formed by 16 lines or by 4 lines?

Question

Is the truth table of $(p↔q)↔(r↔s)$ formed by 16 lines or by 4 lines?

Please help me.

Answer 1

Each of $p$, $q$, $r$, $s$ can take $T$ or $F$. The number of rows is $2^4=16$.

Answer 2

You can't fill the truth table with four lines only because you would be missing some combinations (two, to be precise) of true/false values that statements p, q, r, and s can take.

You must list all possible combinations and their resulting output. To know how many combinations are there, you can use: $$m^n$$ Where m is the number of values a variable can get (in this case, true and false sum up 2), and n is the number of variables (in this case, 4):$$2^4=16$$ You need exactly 16 rows to complete the truth table of your statement.

Maybe what you're trying to say is that only two of those combinations make the statement true, which is correct. But in a truth table, you also want to include the combinations that make the statement false.

Answer 3

In four rows you can do this...

$\def\true{\color{navy}{\mathrm T}}\def\false{\color{blue}{\mathrm F}} \def\getsto{\leftrightarrow} \begin{array}{c:c|c}p\getsto q & r\getsto s & (p\getsto q)\getsto(r\getsto s) \\\hline \true & \true & \true \\ \hdashline \true & \false & \false \\ \hdashline \false & \true & \false \\ \hdashline \false & \false & \true \end{array}$

But that is but a summary, and does not explicitly display all possible assignments for $p,q,r,s$ that generate those evaluations for $p\getsto q$ and $r\getsto s$.

The full table takes $16$ rows.