fa.functional analysis - Question about projections on a Hilbert space

Sorry for the vague title, I can't think of a better one that isn't overly long.

Suppose that $S$ is a commuting set of projection operators on a Hilbert space. I'll introduce the following notation: if $p in S$, let $p^+ equiv p$ and $p^- equiv 1 - p$. Let $I equiv ${$+, -$}. The projections are ordered by defining $p leq q$ whenever the range of $p$ is contained in the range of $q$; this makes the set of all projections into a complete lattice. Is the following identity true?

$sup_{f in I^S} inf_{p in S} p^{f(p)} = 1$

In the case where $S$ is finite with elements $p_1, p_2, ldots p_n$, the left hand side of this equation is simply the product over $i$ of $p_i + (1 - p_i)$, so I'm interested in whether this can be generalised to the infinite case. It's easy to see that the following two statements are equivalent to the above:

If $inf_p p^{f(p)} x = 0$ for all $f$, then $x = 0$

If $sup_p p^{f(p)} x = x$ for all $f$, then $x = 0$

but I have no idea how to prove either of these.

My reason for asking is that I'm trying to show that, if $mathcal{H}_1$ and $mathcal{H}_2$ are Hilbert spaces, then if a projection on $mathcal{H}_1 otimes mathcal{H}_2$ is of the form $sup_i p_i otimes q_i$, with $p_i$ and $q_i$ drawn from some complete Boolean algebras of projections on $mathcal{H}_1$ and $mathcal{H}_2$ respectively, then the $q_i$ may be chosen to satisfy $q_i q_j = 0$ when ever $i neq j$. So if anybody knows of an alternative way to prove that, or knows that it's false, then by all means say so.