Associative Laws

Question

Associative laws: $\begin{align}(p\lor q)\lor r&\equiv p\lor(q\lor r)\\(p\land q)\land r&\equiv p\land(q\land r)\end{align}$

Just curious is $(r\land p)\land q$ also the same as $(p\land q)\land r$? Like are there $3$ options here because I'm only seeing two options.

I mean I do also have commutative laws that say q∧p and p∧q are the same but I'm not sure how that helps prove what I'm asking to prove and I'm still kinda confused here so is what I said right? — jayblosem11, Mar 07 '22 at 18:43

score 1 · Answer 1 · answered Mar 07 '22 at 19:20

1

The identity $(r\land p)\land q = (p\land q)\land r$ holds if $\land$ is commutative and associative, but it may fail if $\land$ is commutative but not associative, as in the example below: $$ \begin{array}{c|rrr|} \land&p&q&r \\ \hline p&p&r&q \\ q&r&q&p \\ r&q&p&r \\ \hline \end{array} $$ In this example, $(r\land p)\land q = q \land q = q$ but $(p\land q)\land r = r \land r = r$.

answered Mar 07 '22 at 19:20

J.-E. Pin

40,163

I honestly don't even understand this answer or any of them.. I don't get this truth table this isn't like any truth tables I've seen before – jayblosem11 Mar 07 '22 at 19:32
I mean isn't ∧ always commutative and associative? When would it not be..? – jayblosem11 Mar 07 '22 at 21:06
Here $\land$ is just a binary operation on the set ${p, q, r}$. – J.-E. Pin Mar 07 '22 at 21:32
So it is associative right? – jayblosem11 Mar 07 '22 at 23:39
No, it is not associative. – J.-E. Pin Mar 08 '22 at 03:50

score 0 · Answer 2 · answered Mar 07 '22 at 18:03

0

That is true if and only if it is conmutative.

Conmutation for operation ∨: $a ∨ b=b ∨ a$

answered Mar 07 '22 at 18:03

Carlos S.G.

33

1

This is not true, see my answer. – J.-E. Pin Mar 07 '22 at 19:21

Associative Laws

2 Answers2