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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?

enter image description here

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  • $\begingroup$ Truth tables are just a convenient tool used for us to organize our thoughts. There is nothing mathematically that says writing it all in one table versus several tables is more or less correct. So long as we understand what we wrote and what it implies that is fine. The difference between writing it in several tables is to me the same difference as writing the ones as green and the zeroes as red. It just doesn't matter and is only personal preference $\endgroup$
    – JMoravitz
    Commented Oct 19, 2016 at 18:48

2 Answers 2

Reset to default
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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.

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  • $\begingroup$ Thank you for answer. Can you please say if the table in edit is correct? Actually it should be same table as yours just that the AND operator is placed at another position in the table, I think. $\endgroup$
    – cnmesr
    Commented Oct 19, 2016 at 18:56
  • $\begingroup$ @cnmesr Yes, the main connective is very clear in your example and it is also easy to compare with the iff truth table. $\endgroup$
    – Adnan
    Commented Oct 19, 2016 at 19:00
  • $\begingroup$ Okay :) I will take table as shown in the edit and also add the equivalent part in it to be bulletproof :P $\endgroup$
    – cnmesr
    Commented Oct 19, 2016 at 19:01
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It may be best to put the whole expression one wants to prove in the truth table.

There are online truth table generators to help avoid errors. Here is the output from one truth table generator showing that the result is valid (a tautology) because the red column under the top level connector symbol is always true (T):

enter image description here

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

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