Let $ \mathbb{P} = \left \{ \bullet , \blacksquare \right \}$ be a set where elements $ \bullet \neq \blacksquare$. For operation of addition in $\mathbb{P}$ we define:
$$\bullet + \bullet = \blacksquare$$ $$\blacksquare + \blacksquare = \blacksquare$$ $$\bullet + \blacksquare= \bullet$$ $$\blacksquare + \bullet = \bullet$$
Verify whether all axioms of addition apply for the set $\mathbb{P}$.
I got stuck showing that $A_2: \forall x,y, z\in \mathbb{P}: x+(y+z)=(x+y)+z$ also works in the set $\mathbb{P}$.
I managed to show that for $x=y=z=\bullet$ and $x=y=z=\blacksquare$ the association works just fine (it required nothing but the use of the definition of the addition (as stated above) plus $A_1 (commutativity)$ which I've managed to prove already. I got stuck at showing that if $$x=\bullet=y \wedge z=\blacksquare $$ then $$\bullet+(\bullet+\blacksquare)=(\bullet+\bullet)+\blacksquare $$
So far my work, starting from the left side of the equation (I used merely the definition of addition in $\mathbb{P}$ and commutativity.
$$\bullet+(\bullet+\blacksquare)=\bullet+\blacksquare+(\bullet+\blacksquare)=\bullet+\blacksquare+\bullet=\bullet+\bullet+\blacksquare$$
so now if I put the $\bullet+\bullet$ in brackets, the problem will be solved. The question is: Can I do that? Can I just add brackets there?
Hilarious how something so trivial can be so puzzling.