I re-read Jechs chapter about forcing, and got a question. There he characterizes a (what he calls) modern way to make the forcing argument legitimate which (I think) goes like this:
It is pointed out there that, in order to establish the consistency of a statement $\varphi$ relative to ZFC, it is sufficient to exhibit a complete Boolean algebra $B$ such that the Boolean value of $\varphi$ in the Boolean-valued model $V^B$ is not zero, i.e. $|| \varphi|| = p \ne 0$.
This fact gives us an alternative approach to forcing, which avoids the assumption of a transitive, countable model $M$ of ZFC, in order to construct the generic extension $M[G]$.
Nevertheless one hardly sees a proof where the Boolean value of an interesting $\varphi$ is really evaluated because it is more easy to pretend that for the p.o. $P \quad$there exists a generic $G$ over $V$, then build $V[G]$ and show that $V[G] \models \varphi$, which is equivalent to the existence of a $p \in G$ such that $p \Vdash \varphi$, which implies that $||\varphi|| \ne 0$.
My problem with this approach is that a generic $G$ over $V$ cannot exist whenever $P$ satisfies the following property: For every $p \in P$ there exist $q \le p, r \le p$ such that $q$ and $r$ are incompatible. (To see this it suffices to realize that for every filter $F$ on $P$ the set {$ p \in P \quad : p \notin F$} is dense, thus a generic $G$ leads to a contradiction). The existence of a generic $G$ does not change the Boolean value of $\varphi$, so I think that the argumentation in the break above remains valid although $G$ cannot exist.
However it seems strange to me that one assumes the existence of thing that must not exist, to faciliate calculation. One can say that this resembles the complex numbers but to me adding a new element to the reals is way weaker than adding an element that actually must not exist to the universe.
So my question is: is the just described strategy of showing the relative consistency of a statement really valid?
