Proof: $(a \leftrightarrow b) \iff (a \rightarrow b) \wedge (b \rightarrow a)$ by using truth table

Question

It makes huge sense that this is true but how to prove it using a truth table? I have started by making a truth table for $(a \leftrightarrow b)$ first:

a      b        a↔b
----------------------
0      0         1
0      1         0
1      0         0
1      1         1

For $a \rightarrow b$ we have:

a      b        a→b
----------------------
0      0         1
0      1         1
1      0         0
1      1         1

And $b \rightarrow a$ we have:

a      b        b→a
----------------------
0      0         1
0      1         0
1      0         1
1      1         1

And now I can see that the second and third lines of the second and third table contradict each other, which means we can just erase them. In the end we have 1 0 0 1 which equals the first table.

How would you solve this task? As I just described? Or would I have to do this $(a \rightarrow b) \wedge (b \rightarrow a)$ ALL in one table? Please do tell me.

Edit: Can I do it like this?

Answer 1

You're missing the last step for the truth table, which is the AND operator:

a→b   ∧   b→a
 1    1    1
 1    0    0
 0    0    1
 1    1    1

      |_ which is the main connective

You can see that the main connective is equal to the truth table for the iff operator.