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
