P v (Q & R) <=> (P v Q) & (P v R)

This is the distributive law of v over &.

Using the rules:

Under P put TTTTFFFF,

Under Q put TTFFTTFF,

Under R put TFTFTFTF,

The rule for "~" (not) is "~T is F and ~F is T",

The rule for "&" (and) is "only T&T is T, all others F",

The rule for "v" (or) is "only FVF is F, all others T",

The rule for "->" (if..then...) is "only T->F is F, all other T",

The rule for "<->" (biconditional "the same") is 
"only T<->T and F<->F are T, all others F,

make this truth table for P v (Q & R) <-> (P v Q) & (P v R)

|P|Q|R|~Q|Q&R|Pv(Q&R)|PvQ|PvR|(PvQ)&(PvR)|Pv(Q&R)<->(PvQ)&(PvR)|
|T|T|T| F| T |  T    | T | T |     T     |        T            |
|T|T|F| F| F |  T    | T | T |     T     |        T            |
|T|F|T| T| F |  T    | T | T |     T     |        T            |
|T|F|F| T| F |  T    | T | T |     T     |        T            |
|F|T|T| F| T |  T    | T | T |     T     |        T            |
|F|T|F| F| F |  F    | T | F |     F     |        T            |
|F|F|T| T| F |  F    | F | T |     F     |        T            |
|F|F|F| T| F |  F    | F | F |     F     |        T            |

The proposition is proved because there are only T's in the last 
column.

Therefore we can replace the biconditional symbol <->, by the
stronger equivalence symbol <=> and write

P v (Q & R) <=> (P v Q) & (P v R)

Edwin