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JavaScript (ES6), 487 467 378 298 292 280 266 264 bytes

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I=&gt;(V=v=&gt;!I[v]||((T=o=&gt;[[]][+!!A[o]]||[(I[v]!=A[o]||A)[o^o&lt;88/8]]+T(++o))(8-8)==I.pop())*V(++v))(V|(A='(){}[]&lt;&gt;\\/ !*+-:=ITX^_|'))//\\(('|_^XTI=:-+*! \//&lt;&gt;[]{}()'=A)|V)((v++)V*(()qoq.I==(8-8)((o++)T+[[8\88&gt;o^o](A||[o]A=![v]I)]||[[o]A!!+][[]]&lt;=o=T))||[v]I!&lt;=v=V)&lt;=I


Defines an anonymous function that takes an array of chars and returns the desired output. Output is truthy/falsy; usually 1/0, but the empty string gives true.

How?

The most obvious trick is to use //\\ as a center point to comment out the mirrored version of the code. After that, it becomes a game of figuring out the shortest way to solve the problem using only the charset given.

The first issue we run into is the lack of keywords and built-ins. We miraculously still have .pop(), but everything else will have to be done via the allowed operators (which includes a[b] and f(c)), with recursion to emulate loops.

The second issue is the lack of logical operators. Neither &amp; and ? are allowed, which means the only decision-making operator we can use is ||. Therefore, we have to carefully structure our logic to account for this.

The first thing I did was to define a function T that mirrors an individual character. The basic idea is to loop through each character in a string of mirror-able chars, testing each for equality with the given char. If it is equal, we return its mirror&mdash;the char at index^1 for (){}[]&lt;&gt;\/, or the char itself for the rest.

The first problem I ran into here was getting either the mirrored char or a falsy value on each iteration. The solution I eventually came up with was (x!=A[o]||A)[o^o&lt;88/8], where x is the input character, A is the mirroring alphabet, and o is the current index. If x is not the same as A[o], this gives true, and the index expression evaluates to undefined; otherwise, the ||A is activated, and we end up getting A[o^(o&lt;11)].

The second problem is how to terminate the recursion. I found that the best way to do this is to simply concatenate the results of every iteration, returning the empty string when the end of A is reached. This presents us with two further problems: converting the undefineds to empty strings, and returning the empty string || something. These can be solved with array abuse: [a]+"" gives the string representation of a, or the empty string if a is undefined. As a bonus, [] is truthy but stringifies to the empty string, so we can conveniently use this as a "truthy empty string".

Now we can use the T function to mirror any single character. We do this recursively, comparing the mirror of I[v++] to I.pop() until the end of the array of chars is reached. We can't use &amp;&amp; or &amp; to check if all of the comparisons are truthy, but so use * instead. Multiplying all of these results together gives 1 if every character is the mirror of the one opposite, or 0 if any comparison fails.

And that is basically how this answer works. I probably didn't explain it very clearly, so please ask any questions you may have and point out any mistakes I have made.