How this might happen is that one branch of your program may have unconditional undefined behavior, which can be detected at the check itself. This would let a compiler elide the entire branch, even side effects that would typically run.
The compiler can elide the unconditional-UB branch and its side effects, and it can elide the check itself. But it cannot elide the input operation that produces the value which is checked, nor can it elide any side effects before that input operation, unless it can statically prove that no input values can possibly result in the non-UB branch.
That example doesn't contradict LegionMammal978's point though, if I understood correctly. He's saying that the 'time-travel' wouldn't extend to before checking the conditional.