Prove the following for a general binary operation I need some help with the following proof, please.
First, a definition below:


*

*A binary operation $p$ on a set $X$ is a function of two variables, whose values lie in $X$: it assigns to each ordered pair $(x, y$) where $x \in X$ and $y \in X$, a unique element $p(x, y)$ of $X$.

*Often a binary operation is denoted by a special symbol, say $\Delta $, in which case its values are written as
$x \Delta y$ instead of $\Delta (x, y)$.
Prove the following formula for a general binary
operation:
\begin{equation}
    \left[ \forall_{x} \left[x \Delta x = x\right] \right] \implies \left[ 1 \Delta 1 = 1 \right]
\end{equation}
We were instructed to be very specific when it comes to the order of proving formulas. For this reason, I'm adding a definition and an axiom - even though it may seem elementary.


*Def. Given two statements $\alpha$ and $\beta$, the statement \begin{equation}\alpha \implies \beta ,\end{equation} read as "$\alpha \ \mathrm{implies} \ \beta $ ", means \begin{equation}``\alpha \  \mathrm{implies} \ \beta " \end{equation} which is the same as to say \begin{equation}``\mathrm{if} \ \alpha \ \mathrm{then} \ \beta" .\end{equation}

*Axm (Deduction Rules for Implication). From $\alpha$ and $\alpha \implies \beta$ we can conclude $\beta$. The statement $\alpha \implies \beta$ can be concluded after $\beta$ was concluded from $\alpha$.


The proof may seem elementary and obviously true, but I am struggling to understand our professor's layout of proofs. I felt like I had a good understanding of basic proofs in Calc I, but then when we started with Abstract Mathematics, everything changed.
I only have one attempt to show you which is as follows (sorry for the alignment):
\begin{align}
    &1. \quad \left[ \forall_{x} \left[x \Delta x = x\right] \right] \implies \left[ 1 \Delta 1 = 1 \right] \\
    &2. \quad \left[x = 1 \right]\\
    &3. \quad \left[x \Delta x = 1\right] \\
    &4.\quad \left[ (1) \Delta (1) = 1\right] \\
    &5. \quad \left[ \forall_{x} \left[x \Delta x = x\right] \right]  \implies \left[ 1 \Delta 1 = 1 \right]
\end{align}
This proof is wrong according to my professor. The professor is also not responding to my emails so I can't ask him why.
Please excuse if my format does not meet the standards of StackExchange. I am still learning LaTeX.
Any help would be appreciated.

We can prove it by Conditional proof :
1) $\forall x  \ [ x \Delta x = x ]$ --- assumed [a]
2) $1 \Delta 1 = 1$ --- from 1) by rule of Universal instantiation

3) $\forall x  \ [ x \Delta x = x ] \to [1 \Delta 1 = 1]$ --- from 1) and 2), discharging [a].


With a different proof system, we can use the Universal instantiation axiom schema :

$\forall x \ \varphi (x) \to \varphi[x/a]$

to get a one-line proof.


The proof you have provided does not specify the rules of inference used.
You have listed only the propositional rule usually called Modus Pones; in order to complete the proof, you have to use also the rules for managing quantifiers.
