Proving This Tautology?

Rome_Leader

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After some simplification work, I've reduced a lengthy expression, which I won't re-type, to:

((p' & q') or (p & r')) or q or r

Where p' = not(p) and so on.

I'm unsure where to go from here. Expanding it only seems to make it worse, and none of my library of identities/equivalencies have helped. Can anyone point me in the direction of the simplest approach?
 
After some simplification work, I've reduced a lengthy expression, which I won't re-type, to: ((p' & q') or (p & r')) or q or r

Have a long look at this webpage,

You posted this: \(\displaystyle \left( {\neg p \wedge \neg q} \right) \vee \left( {p \wedge \neg r} \right) \vee q \vee r\)

By combining the first and third disjunct we get \(\displaystyle \left( {\neg p \vee q} \right)\)

Then the second with the fourth gives \(\displaystyle \left( {p \vee r} \right)\)

But \(\displaystyle \left( {\neg p \vee p} \right) \vee \left( {q \vee r} \right)\) is always true.
 
I understand the truth table construction, and could do that in a pinch, but I am trying to hammer it out by Boolean algebra alone. I see I have complicated my grouping a little, and I follow you to a point, but I have a further question, if I may:

After the disjunction combining, I am left with: (p' or q) or p or r. My question is, what now? In trying to expand, I paired the p with the bracketed expression, and got (p' or p) & (q or p) or r. Obviously (p' or p) is always true, but that leaves me then with q or p or r, which surely cannot simplify further?

EDIT: Disregard that, my grouping has fooled me again. Sometimes I tend to see things in brackets as ANDed when they are not, and taking precedence where they do not. I understand now, and will work on that gap in my understanding.
 
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