my lecturer published a question (and the solution) in logics, and i didn't understood his solution, I will be glad if someone may enlighten me why this is true. Q:
let $X_{B,F}$ the inductive set such that: $$\mathbb{X} = \left\{f\mid f\colon\mathrm{\mathbb{R}\to \mathbb{R}} \right\} \\ B=\left\{ p,q \right\} \\ p(x)=\sin(x),\ \ q(x)=\cos(x) \\ F_1 \colon \mathbb{X^2}\to \mathbb{X},\ \ F_1(f,g)(x) = f(x)+g(x) \\ F_2 \colon \mathbb{X^2}\to \mathbb{X},\ \ F_2(f,g)(x) = f(x)\cdot g(x) \\ F=\left\{ F_1,F_2 \right\}$$
Show that the $0$ constant function and the function $\cos(2x)$ is not in $X_{B,F}$
The answer that published is:
we will prove with structural induction that for any $f\in X_{B,F}$ holds $f(\frac{\pi}{4}) > 0$
I didn't understood why if we prove that claim, then this answers the question? i.e it's true for this specific value of $x$, but if we insert another value of $x$, it doesn't necessarily satisfy what we have been asked for.
base: $p(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}, q(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
step: if $f(\frac{\pi}{4}) > 0$ and also $g(\frac{\pi}{4}) > 0$ therefore:
$F_1(f,g)(\frac{\pi}{4}) = f(\frac{\pi}{4}) + g(\frac{\pi}{4}) > 0\\ F_2(f,g)(\frac{\pi}{4}) = f(\frac{\pi}{4}) \cdot g(\frac{\pi}{4}) > 0$
notice that the constant $0$ function, and $\cos(2x)$ function in - $\frac{\pi}{4}$ equals $0$ and therefore $\notin X_{B,F}$
Here the proof ends, and basically I didn't understood what I have mentioned in the Blockquote.
and how come that $0$ not in $X_{B,F}$? I thinking about that and if we take $p(0) = \sin(0) = 0$
it doesn't contradicts this whole thing?
Two functions are equal if and only if they agree on all values of their domain. Your set $X_{B,F}$ is a set of functions, and so it's not enough to just consider the value that some function takes at one particular value to then conclude it is in the set (but you can use this to possibly decide that a function is not in the set).
The zero function $0$ is not in $X_{B,F}$ because it cannot inductively be formed by taking sums and products of the functions $\sin(x)$ and $\cos(x)$.
That is, you can't write $0(x)$ as something like $$((\sin(x)\sin(x)\cos(x))+\sin(x)\cos(x))(\cos(x)+\sin(x)\cos(x))$$
You've given the proof for why that can't be. Because $0(\pi/4) =0$, but any expression of the above form must take a positive value at the input $\pi/4$. Likewise for the function $\cos(2x)$.