# How can I indicate a truth table if its Valid or Invalid?

Construct a truth table for Destructive Dilemma using the general symbolic notation for the rule of inference, T for true value, F for false value. Indicate whether valid or invalid.

Is this the correct way of proving it?

• It turns out Wikipedia has a sample proof that does not use a truth table. You may find it to be more informative. – Daniel W. Farlow Mar 8 '15 at 6:24
• Do you really want to prove this using a truth table for some reason? If so, why? – Daniel W. Farlow Mar 8 '15 at 6:31

Look at the truth values of the columns given by $(p\to q)$, $(r\to s)$,$(\neg q\lor \neg s)$, and $(\neg p \lor \neg r)$.
Your premises are: $(p\to q)$, $(r\to s)$, and $(\neg q\lor \neg s)$. Check for the rows where each of these premises are true along with the conclusion $(\neg p \lor \neg r)$. If a single row has each of the premises true, but the conclusion false, it is an invalid argument; otherwise, it is a valid argument.
Note that the following is what you are ultimately trying to prove (i.e., the destructive dilemma): $$[(P\to Q)\land(R\to S)\land(\neg Q\lor\neg S)]\implies \neg P\lor\neg R.\tag{1}$$ I will present a truth table solution where we use the following notation to make things more manageable: $$\Omega : (P\to Q)\land(R\to S)\land(\neg Q\lor\neg S).$$ Observe the following: $$\boxed{ \begin{array}{c|c|c|c|c|c|c|c|c|c} P & Q & R & S & P\to Q & R\to S & \neg Q\lor\neg S & \Omega & \neg P\lor\neg R & \color{red}{\Omega\to \neg P\lor\neg R} \\ \hline T & T & T & T & T & T & F & F & F & \color{red}{T} \\ T & T & T & F & T & F & T & F & F & \color{red}{T} \\ T & T & F & T & T & T & F & F & T & \color{red}{T} \\ T & T & F & F & T & T & T & T & T & \color{red}{T} \\ T & F & T & T & F & T & T & F & F & \color{red}{T} \\ T & F & T & F & F & F & T & F & F & \color{red}{T} \\ T & F & F & T & F & T & T & F & T & \color{red}{T} \\ T & F & F & F & F & T & T & F & T & \color{red}{T} \\ F & T & T & T & T & T & F & F & T & \color{red}{T} \\ F & T & T & F & T & F & T & F & T & \color{red}{T} \\ F & T & F & T & T & T & F & F & T & \color{red}{T} \\ F & T & F & F & T & T & T & T & T & \color{red}{T} \\ F & F & T & T & T & T & T & T & T & \color{red}{T} \\ F & F & T & F & T & F & T & F & T & \color{red}{T} \\ F & F & F & T & T & T & T & T & T & \color{red}{T} \\ F & F & F & F & T & T & T & T & T & \color{red}{T} \end{array}}$$ As you can see, what we wanted to prove (what is highlighted in red) is a tautology. This proves the equivalence in $(1)$ using a truth table approach, but I would recommend using a truth table approach as only your last resort. Using deductions is much more elegant!