Recent content by Florence

I think that I get it! I will try it out! Thank you!

No problem! I really appreciate all of your help! But I don't really get what I need to do now Do I need to start the hypothetical statement with ##(p \equiv q)## Then start another hypothetical statement with ## ¬((p∧q)∨(¬p∧¬q))## This? Or ##¬((p∨¬q)∧(¬p∨q))## This?

No we haven't seen that, I'm only in my first year psychology and I think we will only see the basics of logics

So I've set up some truth tables and I was able to find out that I can prove them. But now I am trying an other way. I'm trying to prove ##¬((p∧q)∨(¬p∧¬q)) \to ¬(p \equiv q)## So I started my hypothetical statement with ##¬((p∧q)∨(¬p∧¬q))## ##¬(p∧q)∧¬(¬p∧¬q)## Then I started a new hypothetical...

I think that I do understand what you mean and I see that introducing the negation on ##¬(p∧q)∨(¬p∧¬q)## By doing ##¬((p∧q)∨(¬p∧¬q))→(p≡q)## ##¬((p∧q)∨(¬p∧¬q))→¬(p≡q)## Is unnecessary because when I use the transposition rule on ##¬((p∧q)∨(¬p∧¬q))→¬(p≡q)## I get ##(p \equiv q) \to...

I think that we haven't seen the distributive law in logics. I've put all the rules that we have just below. For this exercise we are not allowed to use truth tables, so we have to use these rules to prove it. ##A, B / A∧B ## ##A∧B / A resp. A∧B / B## ##A / A∨B resp. B / A∨B## ##A∨B, A \to C...

Homework Statement I have to prove that ##(p \equiv q) \equiv ((p ∧ q) ∨ (¬p ∧ ¬q))## With no premisses In order to prove this, I first need to prove that: ##(p \equiv q) \to ((p ∧ q) ∨ (¬p ∧ ¬q))## And: ##((p ∧ q) ∨ (¬p ∧ ¬q)) \to (p \equiv q)## I was able to find the second implication...

Thank you!

Hello everybody! My name is Florence, and I am a psychology student. I hope to learn more about logics on this forum!