For the exam that I am taking, propositional always comes up with identical questions. These include writing a sentences in propositional logic, which I can do. But also drawing a truth table for propositional logic, which I can't do. I find It extremely difficult.

For example, the question

Draw the truth table for the following propositional formula: enter image description here

I understand the truth tables. And is only true when both p and q are true, or is only false when both P and Q are false. Not is opposites so if false then true, if true then false.

However, when it comes to applying these to e.g. the question above, i'm lost from the very beginning. It's pretty important I understand this because it's worth 5 marks = )

  • 1
    $\begingroup$ Here's one example of making a truth table that I did a while back... Can you follow how I made this one? math.stackexchange.com/questions/379680/… $\endgroup$ – apnorton Jan 9 '14 at 13:52
  • $\begingroup$ Not really tbhh $\endgroup$ – user119325 Jan 9 '14 at 14:10
  • $\begingroup$ Do you at least know how to find how many rows the associated truth table will have? $\endgroup$ – Dan Rust Jan 9 '14 at 14:14
  • $\begingroup$ You also missed the description of one basic operator "if...then" or "implies". $p \implies q$ is only false if $p$ is true and $q$ is false. Although, I suppose you already know that... $\endgroup$ – lsoranco Jan 9 '14 at 14:35
  • 1
    $\begingroup$ If the $p\wedge q$ is in brackets, so it's $¬(p\wedge q)$ then you first evaluate $p\wedge q$ and then evaluate the negation of this. $\endgroup$ – Dan Rust Jan 9 '14 at 14:37

So we are supposed to make the truth table for the following propositional formula: $$(p\lor \lnot q )\Rightarrow(q\land r).$$ The first observation is that we'll need $2^3$ rows (we have $p,q$ and $r$). The second is that we'll have to determine also the possible truth values for $\lnot q$, then for $(p\lor\lnot q)$ then for $(q\land r)$, and then we can finally conclude the possible truth values for our original proposition. $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \color{green}{p} & \color{green}{q} & \color{green}{r} & \lnot q & (p\lor\lnot q)& (q\land r) &(p\lor \lnot q )\Rightarrow(q\land r) \\ \hline \color{royalblue}{1} & \color{royalblue}{1} & \color{royalblue}{1} & \color{#C00}{0} & \color{royalblue}{1}& \color{royalblue}{1}&\color{royalblue}{1} \\ \hline \color{royalblue}{1} & \color{royalblue}{1} & \color{#C00}{0} & \color{#C00}{0} & \color{royalblue}{1}& \color{#C00}{0}&\color{#C00}{0} \\ \hline \color{royalblue}{1} & \color{#C00}{0} & \color{royalblue}{1} & \color{royalblue}{1} & \color{royalblue}{1}& \color{royalblue}{1} &\color{#C00}{0} \\ \hline \color{royalblue}{1} & \color{#C00}{0} & \color{#C00}{0} & \color{royalblue}{1} & \color{royalblue}{1}& \color{#C00}{0}&\color{#C00}{0} \\ \hline \color{#C00}{0} & \color{royalblue}{1} & \color{royalblue}{1} & \color{#C00}{0} & \color{#C00}{0}& \color{royalblue}{1}&\color{royalblue}{1} \\ \hline \color{#C00}{0} & \color{royalblue}{1} & \color{#C00}{0} & \color{#C00}{0} & \color{#C00}{0}& \color{#C00}{0}&\color{royalblue}{1} \\ \hline \color{#C00}{0} & \color{#C00}{0} & \color{royalblue}{1} & \color{royalblue}{1} & \color{royalblue}{1}& \color{#C00}{0}&\color{#C00}{0} \\ \hline \color{#C00}{0} & \color{#C00}{0} & \color{#C00}{0} & \color{royalblue}{1} & \color{royalblue}{1}& \color{#C00}{0}&\color{#C00}{0} \\ \hline \end{array}$$


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