How Do I Prove This Tautology?

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AI Thread Summary
The discussion centers on proving the tautology [(p or r) & (not(p) or r)] ---> r without using truth tables. The original poster expresses difficulty in simplifying the expression and seeks a more straightforward method. They initially struggle with the complexity of expanding terms but later find that distributing the implication leads to a quicker solution. Ultimately, they successfully solve the problem and share their approach. The thread highlights the importance of exploring different methods in logical proofs.
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Homework Statement



[ (p or r) & (not(p) or r) ] ---> r is a tautology. Prove this without using truth tables.

Homework Equations



See above. not(p) = !p = p' = the opposite value of p and ---> is an implication.

The Attempt at a Solution



I have made some prior simplification, and that is what I have at present. I don't know where to go from here. Expanding the terms seems to make things too complicated. Is there a shorter way?

Any help would be greatly appreciated!
 
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Nevermind, I've solved it! For anyone interested in the solution, I just had to distribute the implication and then expanding was a lot quicker!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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