# 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!

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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.

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.

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)
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4. p (assumption)
5. p (copy rule)
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1. p ( ∨e 1, 2-3, 4-5) (or elimination)

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1. (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.