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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?

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


An argument is valid if every argument of the same form has premises that are all true along with a true conclusion. An argument is invalid if there is an argument of the form whose premises are true, but the conclusion false.

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  • $\begingroup$ is it not the same as "(P⊃Q)∙(R⊃S); ~Q V ~S/∴ ~P V ~ R"? $\endgroup$ – Natty Guurl Mar 8 '15 at 6:17
  • $\begingroup$ oh nvm u meant constructive $\endgroup$ – Natty Guurl Mar 8 '15 at 6:18
  • $\begingroup$ @NattyGuurl Sorry, I edited it now, it's the same as what you have, just different notation. $\endgroup$ – user144809 Mar 8 '15 at 6:21
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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!

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