How is (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ (p ∧ ¬r) ∨ (¬q ∧ q)? Is it really distributive property? The distributive property is very simple and it says p ∧ ( q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r ), but here how is (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ (p ∧ ¬r) ∨ (¬q ∧ q) which someone told me is according to the distributive property, but I didn't get it.  
In simple words, can someone please tell me in parallel & exactly how this proposition (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ (p ∧ ¬r) ∨ (¬q ∧ q) matches the distributive property p ∧ ( q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r )? What each variable in distributive property means in that proposition. Million Thanks!  
 A: In
(p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ (p ∧ ¬r) ∨ (¬q ∧ q),
let
s = p ∧ ¬r.
Then this becomes
(s ∧ q) ∨ (s ∧ ¬q) ≡ s ∨ (¬q ∧ q),
for which the distributive
property is clear.
Of course
you also need the
commutative and associative properties.
A: Do it step by step, i.e. $(A\land B\land C)\lor(D\land E)\equiv((A\land B\land C)\lor D)\land((A\land B\land C)\lor E)$, etc... 
Notice by example that 
$$(p\land q\land\lnot r)\lor p\equiv p$$
so the expression simplifies a lot in each step.
A: Consider the theorem:  
|- (p ∧ q ∧ ¬r) ∨ p -> p 
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1.(p ∧ q ∧ ¬r) ∨ p (assumption)
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2.(p ∧ q ∧ ¬r)  (assumption)
3. p     (and eimination)
_________________________
_________________________
4. p (assumption)
5. p (copy rule)
_________________________


*p  ( ∨e 1, 2-3, 4-5) (or elimination) 
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*(p ∧ q ∧ ¬r) ∨ p -> p


This is how you can prove it.
Also, in my humble opinion, i don't think anything is getting redistributed here, no at least conventionally. It's just a basic tautology/theorem which can be used to simplify the main expression. 
