> This is true when using compilers as well, right? No, because ... > See Reflections on Trusting Trust" by Ken Thompson That still results in `f(x) -> y`, n

> This is true when using compilers as well, right? No, because ... > See Reflections on Trusting Trust" by Ken Thompson That still results in `f(x) -> y`, not `f(x) -> P(y)`.