I'm going through "Introduction to Lattices and Order" by Davey and Priestley (2nd ed.) and I'm struggling with a detail in the proof of the $M_3$-$N_5$ theorem. I understand every part of the proof except for one detail. In p. 91, in the proof that $u \wedge v = p$ I can't figure out how it goes from line 3 to 4. Reproducing these first four lines here:
\begin{align} u \wedge v & = ((d \wedge q) \lor p) \wedge ((e \wedge q) \lor p) \\ & = ((e \wedge q) \lor p) \wedge (d \wedge q)) \lor p) & \textrm{by (M)} \\ & = ((q \wedge (e \lor p)) \wedge (d \wedge q)) \lor p) & \textrm{by (M)} \\ & = ((e \lor p) \wedge (d \wedge q)) \lor p & \textrm{by ???} \\ & = \cdots \end{align}
It's the last one that I don't get. It requires that $q \wedge (e \lor p) = (e \lor p)$, but how? Certainly this would follow from the Connecting Lemma if we knew that $e \lor p \le q$, but I don't see how this is the case.
For reference, it is assumed that the lattice is modular but not distributive, and that $(d \wedge e) \vee (d \wedge f) < d \wedge (e \vee f)$. And the following are defined:
\begin{align} p & := (d \wedge e) \vee (e \wedge f) \vee (f \wedge d) \\ q & := (d \vee e) \wedge (e \vee f) \wedge (f \vee d) \\ u & := (d \wedge q) \vee p \\ v & := (e \wedge q) \vee p \\ w & := (f \wedge q) \vee p \\ \end{align}
I have tried a few things:
We know that $p \le q$, so it would be enough to show that $e \le q$ to obtain that $e \lor p \le q$, but how can we prove that $e \le q$?
Alternatively, since $q = \wedge \{d \vee e, e \vee f, f \vee d\}$, if we show that $e \vee p$ is a lower-bound of $\{ d \vee e, e \vee f, f \vee d \}$ then $e \lor p \le q$ would follow. But while it is easy to show that $e \vee p \le d \vee e$ and that $e \vee p \le e \vee f$, I don't see how to show that $e \vee p \le f \vee d$.