I'm having trouble proving this identity found in Curry's Foundations of Mathematical Logic.
Let $L$ be a lattice. For all $a,b,c,d \in L$, show that: $$(a \wedge b) \vee (c \wedge d) \leq (a \vee c) \wedge (b \vee d) $$
I've tried to first show: $$a \wedge b \leq (a \vee c) \wedge (b \vee d)$$ $$ c \wedge d \leq (a \vee c) \wedge (b \vee d)$$ but I'm getting nowhere.
Is this a good approach, or should I try something different? Should I try finding a counter-example?
$$a\land b\le a\le a\lor c$$ $$a\land b\le b\le b\lor d$$ $$a\land b\le a\lor c\quad\&\quad a\land b\le b\lor d\implies a\land b\le(a\lor c)\land(b\lor d)$$