Let $L$ be a lattice and $x,y,z\in L$.
If $y \leq z$, then clearly $x\vee y \leq x\vee z$ and $x\wedge y \leq x\wedge z$.
Now I wonder about the reverse direction. In general, $x\vee y \leq x\vee z$ does not imply $y \leq z$ (take for example $x = 1$ in a lattice with 1). Neither does $x\wedge y \leq x\wedge z$.
But what about taking both those conditions together? Does $x \vee y \leq x\vee z$ and $x \wedge y \leq x\wedge z$ together imply $y\leq z$?
My feeling is that the answer should be "yes", but so far I've failed to rigorously prove it.
If necessary, we may assume that the lattice under consideration is modular.
The answer is no. For a counterexample, take $N_5 = \{a,b,b',c,d\}$, with $a<b<b'<d$, and $a<c<d$. We have $d = c \vee b' \leq c \vee b = d$, and $a = c \wedge b' \leq c \wedge b = a$, but $b' > b$.
Actually the condition you stated seems to be equivalent to distributivity (and therefore imply modularity). $N_5$ isn't distributive, hence the counterexample.