I'm trying to use the De Morgan and other rules of simplification but I'm going nowhere. Is this supposed to be hard or am I missing something?
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First, use commutivity and association:
$${\quad p\vee\big((p\vee \neg q)\wedge(q\vee \neg r)\big)\vee\neg q\vee r\\=(p\vee\neg q\vee r)\vee\big((p\vee\neg q)\wedge(q\vee\neg r)\big)\\=}$$
Next use distribution of $\vee$ over $\wedge$.
Then it shall become clear what to do to polish it off.
Graham Kemp
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Using Boolean algebra,
$$\begin{align} p + (p + \bar q) \cdot (q + \bar r) + \bar q + r &= p + p q + p \bar r + \overbrace{\bar q q}^{=0} + \bar q \bar r + \bar q + r\\\\ &=p \underbrace{(1 + q + \bar r)}_{=1} + 0 + \bar q \underbrace{(\bar r + 1)}_{=1} + r = \color{blue}{p + \bar q + r}\end{align}$$