0

I don't know how I could answer the truth table of $(p \lor q \lor ¬r) \land (p \lor ¬q \lor ¬s) $ because it's kind of confusing because it has 3 variables in one parentheses and I want some specific and clarified answer.

user585627
  • 1
  • 1
    Well, I recommend determining the truth tables of $p\vee q\vee\neg r$ and $p\vee\neg q\vee\neg s,$ first. Then, you'll basically be done. – Cameron Buie Aug 20 '18 at 02:20
  • Since disjunction is associative, there's no need to add parentheses. $p\lor q \lor \neg r \equiv (p \lor q) \lor \neg r \equiv p\lor (q\lor \neg r)$ – eyeballfrog Aug 20 '18 at 02:24
  • how would i construct the truth table – user585627 Aug 20 '18 at 02:44
  • 1
    @user585627 how many rows have the truth table? – manooooh Aug 20 '18 at 03:18
  • @eyeballfrog No, that doesn't work in this context, since p∨q∨¬r doesn't refer to a unique well-formed formula. In terms of being well-formed formulas, ((p∨q)∨¬r) is distinct from (p∨(q∨¬r)). The truth tables for ((p∨q)∨¬r) and (p∨(q∨¬r)) are distinct in that one has (p∨q) computed, while the other has (q∨¬r) computed. – Doug Spoonwood Aug 20 '18 at 21:55
  • @DougSpoonwood Is there an assignment of truth values such that $((p\lor q)\lor \neg r)$ is true and $(p \lor (q\lor\neg r))$ is false (or vice versa)? If not, I don't get why the distinction you draw matters. – eyeballfrog Aug 20 '18 at 21:59
  • @eyeballfrog No, there is not. But, there is no assignment of truth values such that (p$\rightarrow$p) is true and (p$\rightarrow$(q$\rightarrow$p)) is false also. So, just because we have equivalent truth values, that doesn't imply that no distinction exists. Also, consider how to prove ((p∨q)∨¬r) vs. how to prove (p∨(q∨¬r)). – Doug Spoonwood Aug 21 '18 at 00:39
  • List out all values for every list of (p, q, r, s) first. A list such as (0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0), ..., (1, 1, 1, 1) Then write out the truth tables for every subformula of your intended formula. – Doug Spoonwood Aug 21 '18 at 00:42

1 Answers1

0

Because disjunction is associative you have $(p \vee q ) \vee \lnot r=p \vee (q \vee \lnot r)$. To construct the truth table for this you would have $2^3=8$ lines, one for each of the possible combinations of truth values of $p,q,r$. Because of the associativity you can make a column with $p \vee q$ and then with $(p \vee q) \vee \lnot r$. You should find seven true values and one false value. When you add in the other conjunct you need $2^4$ lines because you have a fourth variable.

Ross Millikan
  • 374,822