Understanding Solving Simple Equations I'm having trouble explaining/understanding how the theorem 

(1) $\forall x,y,z\in\mathbb R(x=y\iff x+z=y+z)$  

applies to solving simple equations. Suppose we are working with the equation $p_1(x): x+1=0$ and we wish to determine the values of $x$ for which $p_1(x)$ is true. Rather than carrying out the usual "add $-1$ to both sides..." I want to understand why (1) says that $p_1(x)$ is equivalent to $p_2(x): x=-1.$
The theorems says:

$x +1 = 0 \iff x + 1 + z = 0 + z$ for all $z$.

In particular $x+1 = 0 \iff x+1 + (-1) = 0 + (-1)$

So $x+1 = 0 \iff x = -1$.

The theorem says if you ever come across $x + a = c$ you can solve it by adding $-a$ to both sides because $x + a = c \iff x + a +(-a) = c +(-a)$.
Well, let's use your example. Say we got the formula $p_1(x) : x + 1 = 0$. We state $x = y$, Just like you said. Since we got $x + 1$ on the left hand side, it must be that $z=1$. Now given your definition of $x$, $y$ and $z$, $x = y$ and $x+z=y+z$ are interchangeable. That means that if we get rid of $z$ on both sides, it states $x=y$. $$x + z = 0$$
$$y + z = 0$$
$$y = -z$$
$$x = y = -z$$
Since $z=1$ $x=y=-1$.
Just as clarification, there is no real use in using the way we just did. We better could've said $x=0-1=-1$ at the beginning. One of the explanations of this is that, not only is the formula clearly just another way of writing this, it also only has 1 unknown (or 2 of $z$ is unknown too). So I hope I made clear what the theorem does, but basically it's still the same.