# $P$: $((p \land q) \vee r)\implies (l \vee t)$… If $p$,$q$,and $l$ are all false and $r$ and $t$ are true determine if $P$ is true

$P$: $((p \land q) \vee r)\implies (l \vee t)$... If $p$,$q$,and $l$ are all false and $r$ and $t$ are true determine if $P$ is true.

How would I show this just by logically writing out $p$ and $q$ false or $r$ true implies $l$ false or $t$ true so $P$ is true? Or would I have to make a truth table which would be complicated. $P$ is true right?

• The title isn't supposed to be the first line of your question. – Git Gud Apr 30 '14 at 23:37
• Its not a title, if I just have the second part you would not know what the problem was and visa versa. – user146167 Apr 30 '14 at 23:40
• Exactly. I just look at the question (I just don't read "titles" once I open the question) and I can't make head or tails with what it is. It doesn't make sense. – Git Gud Apr 30 '14 at 23:42
• Well I don't know what type of problem it is to call it anything different, and the way I look at filling out the boxes is question on the top box and then any addition thoughts I have on solving it or questions i have about it. – user146167 Apr 30 '14 at 23:51
• In fact when you go to post a question, the spot for title says what is the question? – user146167 Apr 30 '14 at 23:55

Doing a truth table is one way to prove it. If you don't want to do it you can argue as follows.

Whenever the consequent of a conditional statement is true, the conditional statement is true. Since $t$ is true, so is $l\lor t$, therefore $P$ is true.

• I would have to use that same logic for the other side where p and q false or r true also right? – user146167 May 1 '14 at 0:12
• The LHS of $\implies$ doesn't matter, that's the whole point of the remark "Whenever the consequent of a conditional statement is true, the conditional statement is true". A conditional statement is false only if the antecedent is true and the consequent is false. So if the consequent is true, necessarily the conditional statement is true. Is it clearer? – Git Gud May 1 '14 at 0:13
• Ok so say the whole thing was $a \implies b$ the only time that the proposition is false is where a is true and b is false, i get that but see where a and be and both false the proposition is true ...ahhh i got it you showed that b was true and in both cases of a the proposition was true ....Thanks :) – user146167 May 1 '14 at 0:23
• @user146167 Yes, it doesn't matter what $a$ is. No problem. – Git Gud May 1 '14 at 0:24

Note this: $((p \land q) \vee r) \rightarrow ((F \land F) \vee T) \rightarrow F\vee T\equiv V\\ (l\vee t) \rightarrow (F \vee T) \equiv T\\ T \Rightarrow T \equiv T$

Then $P$ is true.