Algebraic manipulation with $\neq$ instead of $=$. Does ordinary algebraic manipulation work when $=$ is replaced with $\neq$?
For example, is the following a valid statement:
$p-1 \neq 6z\ \Rightarrow p \neq 6z + 1$
And if so, to what extent can this be expanded to other operations?
Edit: I guess formalising my question gave me the intuition to answer it (in the case of the simple example). We can suppose that $p - 1 \neq 6z$. But $p = 6z + 1 \Rightarrow p - 1 = 6z$, and so we have a contradiction. The expansion to other operations part of the question is still a mystery.
 A: It's OK to do this if the action you take on both sides of the equation is invertible. In your example, adding $1$ is an invertible operation.
Adding a fixed number, subtracting a fixed number, multiplying by a nonzero number, and dividing by a nonzero number are all invertible actions you can take. So is raising to a power when that is an invertible action.
A: sometimes.
$p-1 = 6z \iff p = 6z+1$ so $p-1 \ne 6z \iff p \ne 6z+1$ so that is okay.
But
$x = 5 \not \iff x^2= 5^2$ so we can't say $x\ne 5 \implies x^2 \ne 25$.(Because $(-5)^2 = 25$.)
But we CAN so if $x^2 \ne 25 \implies x \ne 5$. (Because $5^2$ can't NOT equal $25$.)
Negation goes in the REVERSE order!
Think of this in terms of logic.  
If $A \implies B$ then $\lnot B \implies \lnot A$.
So if $A = B \iff f(A) = f(B)$ (where $f(A)$ is something like $A+c$ and $m*D$) Then we can also to $f(A) = f(B) \implies A = B$ and $A\ne B \iff f(A) \ne f(B)$.
But if we only have $A=B \implies f(A)=f(B)$ but not $f(A)= f(B) \implies A=B$ (ex.  $f(a) = a^2$ or $f(a) = a^2$ then we can Not go $A \ne B \implies f(A) \ne f(B)$ (notice $x \ne 7 \not \implies |x| = |7|$ as $|-7| = |7|$ but $-7 \ne 7$.  But we CAN go $f(A) \ne f(B) \implies A \ne B$ (Notice!  $|x| \ne |7|$ then $x \ne 7$!!!)
A: Good question.  In general, be very careful when trying this.
In your stated example, it's okay.  Here's why.   If you were wrong and $p=6z+1$ then you are able to deduce $p-1=6z$, which you know to be false.  So, we're good here.
However, be VERY careful trying this with other more complicated things.  For example, the following is fallacious:
$$
x \neq 1 \Rightarrow x^2 \neq 1.
$$
This is wrong, for $x$ could have been $-1$.
In general, make sure that you can justify that this kind of move is right by looking at an equivalent formulation that uses (the much safer) equal signs.
