Proof of equality using basic axioms

Question

I'm supposed to prove equivalence associativity using propositional logic axioms. My teacher insists that I use mathematical symbols. Half of the proof is given and I am to derive the second half.

The start is to prove:

$((p=q)=r) =(p = (q = r))$

The first half was done. Using axioms, $((p=q)=r)$ resolves to:

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

In order to prove that the other side is equivalent, I chose to start with the second side, $(p = (q = r))$, and try to reach the same line as above. Below are my steps, and I can almost see it, but I can't find an axiom that lets me simplify it further without going in another direction that takes me further from where I want to be. In other words, I have no idea if I'm on the right path and if so, what is the next step...

Here's my steps so far:

$\begin{align} (p = (q = r)) = (p\Rightarrow(q = r))\wedge((q=r)\Rightarrow p) \tag{Equality axiom} \end{align} $

Going after $(p\Rightarrow(q = r))$ side first, which will give half of the line from above I'm trying to get to:

$\begin{align} (p\Rightarrow(q = r)) &= (p \Rightarrow(q\Rightarrow r \wedge r\Rightarrow q))\\ &=(p \Rightarrow((\neg q\vee r) \wedge (\neg r\vee q))\\ &=(\neg p \vee ((\neg q \vee r)\wedge(\neg r \vee q)))\\ &=(\neg p \vee (\neg q \vee r))\wedge(\neg p\vee(\neg r \vee q)) \end{align} $

Now that last line looks very close to $(\neg p \vee \neg q \vee r) \wedge (\neg p \vee q \vee \neg r)$ except for the parentheses. Is there an axiom with the disjunction operator that allows me to ignore them? Or did I make a mistake and am I going in the totally wrong direction?

After thinking a bit more and drawing a truth table, I see that $ (p \vee (q \vee r))$ is equivalent to $(p \vee q \vee r)$.

But what is the axiom for this?

Answer 1

Now that last line looks very close to $(¬p∨¬q∨r)∧(¬p∨q∨¬r)$ except for the parentheses. Is there an axiom with the disjunction operator that allows me to ignore them?

Because disjunction is associative you may dispense with the brackets in a dijunctive series without detracting from the meaning of the statement.

$$((\neg p\vee \neg q)\vee r) \\ \equiv\\ (\neg p\vee (\neg q\vee r)) \\ \equiv \\ (\neg p\vee \neg q\vee r)$$

Likewise, because disjunction is commutative you may exchange the order of a disjunctive series and remain equivalent.

$$(¬p∨q∨¬r) \;\equiv\; (¬p∨¬r∨q)\qquad\text{et cetera}$$