Prove that (in a Boolean algebra) if $a + x = b + x$, and $a + x′ = b + x′$, then $a = b$

Question

The proof should be built of the following postulates, for a Boolean Algebra:

P1. The operations (+) and (·) are commutative.

P2. There exist in B distinct identity elements 0 and 1 relative to the operations (+) and (·), respectively.

P3. Each operation is distributive over the other.

P4. For every a in B there exists an element a′ in B such that $a + a′ = 1$ and $aa′ = 0$

More importantly, what is the reasoning for each step?

All I've managed is to go down different paths hoping to eventually see some clues (which hasn't worked) but this can't be an efficient way to approach this

Edit: Please see my attempts in attached photo

Answer 1

I think what you are looking for is:

If $ a + x = b+x $ then $$\begin{align} (a+x)x' &= (b+x)x' \\ ax' +xx' &= bx' + xx' & \text{operator } \cdot \text{ is distributive} \\ ax' &= bx' & \text{because }xx' = 0 \end{align}$$ Similarly, using $a+x' = b+x'$ you obtain $ax = bx$, so that $$\begin{align} ax'+ax &= bx'+bx \\ a(x'+x) &= b(x'+x) & \text{operator } \cdot \text{ is distributive} \\ a\cdot 1 &= b \cdot 1 &x+x' = 1\\ a &= b&1 \text{ is an identity} \end{align}$$

It seems you have not stated the operators are associative nor that the inverse is unique. I suspect that limits the steps you can take.