When does (x == x+2)?

Fortran IV:

2=0

After this every constant 2 in the program is zero. Trust me, I have done this (ok, 25 years ago)

This seems to work:

#define x 2|0

Basically, the expression is expanded to (2|0 == 2|(0+2)). It is a good example of why one should use parentheses when defining macros.

Brainfuck

x

This does of course stretch "evaluate to true" a bit, because in Brainfuck nothing actually evaluates to anything – you only manipulate a tape. But if you now append your expression

x
(x == x+2)

the program is equivalent to

+

(because everything but <>+-[],. is a comment). Which does nothing but increment the value where we are now. The tape is initialised with all zeros, so we end up with a 1 on the cursor position, which means "true": if we now started a conditional section with [], it would enter/loop.