Use inference rules to prove distributive law I'm taking an intro logic course this semester and my prof is hard to follow and not really great at clarifying things. I'm stuck on this question in my assignment, I'm just not sure how to start. I think I just need a jump start on how to tackle this sort of question. 

So the question is, using basic inference rules, provide derivation to prove distributivity of $\lor$ over $\land$: 

$P \lor (Q \land R) \vDash (P \lor Q) \land (P \lor R)$

$(P \lor Q) \land (P \lor R) \vDash P \lor (Q \land R)$
You can prove them with Natural Deduction.

For $P∨(Q∧R) \vDash (P∨Q)∧(P∨R)$ :

1) $P∨(Q∧R)$ --- assumed

2) $P$ --- assumed [a] for $\lor$-elimination (or proof by cases)

3) $P \lor Q$ --- from 2) by $\lor$-introduction

4) $P \lor R$ --- from 3) by $\lor$-introduction

5) $(P∨Q)∧(P∨R)$ --- from 3) and 4) by $\land$-introduction

6) $Q \land R$ --- assumed [b] for $\lor$-elimination

7) $Q$ --- from 6) $\land$-elimination

8) $P \lor Q$ --- from 7) by $\lor$-introduction

9) $R$ --- from 6) $\land$-elimination

10) $P \lor R$ --- from 9) by $\lor$-introduction

11) $(P∨Q)∧(P∨R)$ --- from 8) and 10) by $\land$-introduction

Now, having derived $(P∨Q)∧(P∨R)$ from $P$ (see 2-5) and from $Q \land R$ (see 6-10) we can derive it from $P∨(Q∧R)$ by $\lor$-elimination, discharging assumptions [a] and [b] :


  12) $(P∨Q)∧(P∨R)$ --- from 2), 5), 6) and 11) by $\lor$-elimination.


The above proof shows that :


  
    $P∨(Q∧R) \vdash (P∨Q)∧(P∨R)$;
  


by completeness, we conclude with : $P∨(Q∧R) \vDash (P∨Q)∧(P∨R)$.



In the same way, we can proceed for : $(P∨Q)∧(P∨R) \vDash P∨(Q∧R)$.



You can easily modify the above proof according to this List of rules of inference :


steps 3 and 4 are Addition (or Disjunction Introduction)
step 5 is Adjunction (or Conjunction Introduction)
after step 5 we need Deduction theorem (or Conditional Introduction)
the same set of rules must be used for steps 6-11
finally, we have to apply in step 12 Case analysis.
Proving the Distributive Rule 

x &amp; (y V z) :: (x &amp; y) V (x &amp; z)

In order to prove the Distributive Rule, you must show that BOTH of the following statements are true:

                 Statement #1:     x  &amp; (y V z) → (x &amp; y) V (x &amp; z)   

                                         AND

                   Statement #2:   (x &amp; y) V (x &amp; z) → x  &amp; (y V z)


I show a separate proof for each of these statements:

Proof of Statement #1:    x  &amp; (y V z) → (x &amp; y) V (x &amp; z)

x  &amp; (y V z)                                       P
  x                                                      1 Simp
  y V z                                                1 Simp
  ~ (x &amp; y)                                           AP
  ~ x V ~ y                                           4 DeM
  ~ y                                                    2, 5  DS
  z                                                       3, 6  DS
 x &amp; z                                                  2, 7  Conj
 ~ (x &amp; y) → (x &amp; z)                             4-8  CP
  (x &amp; y) V (x  &amp; z)                             9 DeM   

Proof of Statement #2:   (x &amp; y) V (x &amp; z) → x  &amp; (y V z)

(x &amp; y) V (x &amp; z)                                  P
 ~ x                                                      AP
 ~ x V ~ y                                             2 Add
 ~ x V ~ z                                             2 Add
 ~ ( x &amp; y)                                             3 DeM
 ~ ( x &amp; z)                                             4 DeM
 ~ ( x &amp; y) &amp;  ~ ( x  &amp; z)                        5, 6 Conj
 ~ ( ( x &amp; y) V ( x &amp; z) )                         7 DeM
 ~ (( x&amp;y) V (x&amp;z))  &amp;  ((x&amp;y) V (x&amp;z))  1, 8 Conj
 x                                                          2-9  IP
 ~ ( y V z)                                             AP
 ~  y  &amp; ~ z                                               11 DeM
 ~ y                                                          12 Simp
 ~ z                                                          12 Simp
 ~ z V ~ x                                                 14 Add
 ~ ( x &amp; z)                                                 15 DeM
 x &amp; y                                                        1 DS
 y                                                               17 Simp
 y V z                                                         18 Add
 ~ ( y V z) &amp; ( y V z)                                   11, 19  Conj
 y V z                                                          11 - 20  IP
 x &amp; ( y V z)                                                 10, 21 Conj



We can also prove the other Distributive Rule:

x V (y &amp; z) :: (x V y) &amp; (x V z)

In order to prove this Distributive Rule, you must show that BOTH of the following statements are true:

                 Statement #1:     x  V (y &amp; z) → (x V y) &amp; (x V z)   

                                         AND

                   Statement #2:   (x V y) &amp; (x V z) → x  V (y &amp; z)


I show a separate proof for each of these statements:

Proof of  Statement #1:     x  V (y &amp; z) → (x V y) &amp; (x V z)  

x  V (y &amp; z)                                                                P
~ ( x V y)                                                                   AP
~ x &amp; ~ y                                                                   2 DeM
~ x                                                                            3 Simp
~ y                                                                            3 Simp
y &amp; z                                                                        1, 4 DS
y                                                                               6 Simp
y &amp; ~ y                                                                      5, 7 Conj
 x V y                                                                     2 - 8 IP
 ~ ( x V z)                                                               AP
 ~ x &amp; ~ z                                                               10 DeM
 ~ x                                                                         11 Simp
 ~ z                                                                         11 Simp
 y &amp; z                                                                      1, 12 DS
 z                                                                             14 Simp
 z &amp; ~ z                                                                    13, 15 Conj
 x V z                                                                        10 - 16 IP
 ( x V y ) &amp; ( x V z)                                                    9, 17 Conj

Proof of Statement #2:   (x V y) &amp; (x V z) → x  V (y &amp; z)

( x V y ) &amp; ( x V z )                                                 P
 x V y                                                                      1  Simp
 x V z                                                                      1 Simp
 ~ x                                                                         AP
 y                                                                            2, 4 DS
 z                                                                            3, 4 DS
 y &amp; z                                                                      5, 6 Conj
  ~ x → ( y &amp; z )                                                      4 - 8 CP
  x V ( y &amp; z )                                                           8 Impl