Simplify $(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r)$ Anyone can help me  simplify $$(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r)$$ by using algebraic propositions laws? I'm stuck.
According to a program I'm using the result is: $(\neg p \wedge r) \vee (\neg q \wedge r)$.

Does anyone know a software or webpage that shows you the steps?
$$(\neg p \wedge q \wedge r) \vee \neg(q\vee \neg r) \Leftrightarrow \text{ (DeMorgan)}$$

$$(\neg p \wedge q \wedge r) \vee (\neg q \land \neg \neg r) \Leftrightarrow \text{ (Double Negation)}$$

$$(\neg p \wedge q \wedge r) \vee (\neg q \land r) \Leftrightarrow \text{ (Distribution)}$$

$$((\neg p \wedge q ) \vee \neg q) \land r \Leftrightarrow \text{ (Distribution)}$$

$$((\neg p \vee \neg q) \wedge (q  \vee \neg q)) \land r \Leftrightarrow \text{ (Complement)}$$

$$((\neg p \vee \neg q) \wedge \top) \land r \Leftrightarrow \text{ (Identity)}$$

$$(\neg p \vee \neg q) \land r \Leftrightarrow \text{ (Distribution)}$$

$$(\neg p  \land r) \vee (\neg q \land r) $$

There are some advanced equivalence principles that can do this a bit more quickly. For example, if you have:

Reduction

$p \land (\neg p \lor q) \Leftrightarrow p \land q$

$p \lor (\neg p \land q) \Leftrightarrow p \lor q$

then you could go from line 4 immediately to line 7

In fact, the generalized form of Reduction says:

Reduction (generalized)

$(r \lor p) \land (r \lor \neg p \lor q) \Leftrightarrow (r \lor p) \land (r \lor q)$

$(r \land p) \lor (r \land \neg p \land q) \Leftrightarrow (r \land p) \lor (r \land q)$

and with that you can go from line 3 to line 8 in 1 step!
HINT: Check this:

Wolfram Alpha

https://www.wolframalpha.com/input/?i=(not+p+AND+q+AND+r)+OR+not+(q+OR+not+r)