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How would someone go about simplifying the expression:

$((p \vee r)\wedge((p \wedge q) \vee (p \wedge ¬q)))\vee (p \wedge r) \vee \neg(p \vee q) \vee r$

Any help would be appreciated.

EDITED: Is this a correct truth table?

enter image description here

barak manos
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Ebu Potokombe
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  • Start with a truth-table, in order to illustrate to yourself the $4$ possible cases of this expression. And while you're at it, make sure that you illustrate it to the rest of us as well (i.e., show your effort if you're expecting others to make an effort for you). – barak manos Dec 04 '16 at 22:32
  • Okay, can you please tell me what specific boolean laws are there, because I type boolean laws on youtube it comes up with different kind of boolean laws i.e: AB'C'D + ABC' etcetc. – Ebu Potokombe Dec 04 '16 at 22:37
  • Sorry, I just noticed the $r$, so there are 8 possible cases. – barak manos Dec 04 '16 at 22:38
  • You really need to add some parentheses! – Bram28 Dec 04 '16 at 23:19
  • Barak, could you check my truth table, please? – Ebu Potokombe Dec 05 '16 at 01:17
  • @EbuPotokombe Ok, I will add where I believe parentheses should go. If I am wrong, then you can change it again, but again, you really need some parentheses, because otherwise your expression is not grammatical and we cannot be sure what it is. – Bram28 Dec 05 '16 at 03:17
  • No, your truth table is wrong. You didn't even get simple stuff like $p\wedge q$ correctly (at least the last line). In addition, you've recklessly replaced $\neg(p \vee q)$ with $p \vee q$. More generally, you've replaced the order of the inner expressions, which makes it hard to compute the final expression. By working in a non-orderly-fashioned manner, you've made your entire effort prone to errors, and ended up with a bunch of silly mistakes. Fix up the table, write it down properly and carefully, then I will be willing to give it a second look. – barak manos Dec 05 '16 at 07:40

1 Answers1

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Hints:

  • $(p \land q) \lor (p \land \lnot q) = p \land(q \lor \lnot q) = p\;$ by distributivity and complements;

  • $(p \land r) \lor r = r\;$ by absorption;

  • $\lnot(p \lor q) = \lnot p \land \lnot q$ by De Morgan's.

[ EDIT ]  After effecting the simplifications above, the expression becomes:

$$((p \lor r) \land p) \lor r \lor (\lnot p \land \lnot q)$$

More hints:

  • $(p \lor r) \land p = p\;$ again by absorption;

  • $p \lor (\lnot p \land \lnot q) = (p \lor \lnot p) \land (p \lor \lnot q)=p \lor \lnot q\;$ by distributivity and complements.

dxiv
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  • @EbuPotokombe Yes, that's what I am getting. You can doublecheck that with the truth table, of course. – dxiv Dec 05 '16 at 01:59
  • ((p∨r∨(¬p∧¬q)((p∨r)∧p)∨r∨(¬p∧¬q) and then to pV¬q ((p∨r)∧p)∨r∨(¬p∧¬q) using distrubivity, absorption and compliments, right? – Ebu Potokombe Dec 05 '16 at 02:06
  • @EbuPotokombe You should really use MathJax since it's difficult to read without it. Associativity is also used, which allows reordering the "$\lor$" disjunction terms. – dxiv Dec 05 '16 at 02:12
  • is the simplified expression "p v ¬q v r" ? – Ebu Potokombe Dec 05 '16 at 02:54
  • @EbuPotokombe I answered that when you first asked an hour ago, before you deleted the previous comment. – dxiv Dec 05 '16 at 03:01
  • Can you answer again, please? Because I can't see what you said as an answer before – Ebu Potokombe Dec 05 '16 at 03:34
  • @EbuPotokombe That's the first comment now, since you deleted yours. "Yes, that's what I am getting. You can doublecheck that with the truth table, of course.". – dxiv Dec 05 '16 at 03:37