How to prove $p∧(q∨r) ⇒ (p∧q)∨(p∧r)$ in logic way not truth table? with the help of equivalence rules.

Question

we have an assignment to prove this distribute law. I can understand this but don't know how to prove it. The teacher left some hint, use some equivalence we had already proved
$$((p∧q)\implies r) \iff (p\implies(\lnot q)∨r)\\ (p∨q)∧\lnot q \implies p\\ p∧\lnot(p∧q)\implies\lnot q$$

Answer 1

This is a rather confusing question, as it is not clear what method you have to or are allowed to use and what logic rules you can and cannot use.

First of all, you say you have to use equivalence rules (which only show things of the form $\Leftrightarrow$) and yet you have to show an implication ($\Rightarrow$). Now of course an implication follows from an equivalence, and as it so happens, what you need to prove is an equivalence, and not just an implication. That is, it is true that:

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

from which the desired implication

$p\land(q\lor r) \Rightarrow (p\land q)\lor (p\land r)$

immediately follows.... but is that what you are supposed to (or allowed to) do?

Even more confusingly, the last two hints are not equivalences ... so are you allowed to use things other than equivalences? If you can only use equivalences, then only the first hint can be used, but that one cannot be applied to either side of your goal, so you would need to introduce some other equivalences.

In fact, even if you can use things other than equivalences (e.g. hints 2 and 3), then it is still not clear how you would get to your goal without some further principles, since none of them can be applied to either side of your goal.

So: can you check the problem statement given by your professor and see 1) what exactly you need to show, and 2) what you can and cannot use?

ADDITION

OK, I still don't see how you could possibly do this with just the hints provided, or even with some further (basic) equivalence principles. I am also still a bit mystified with regard to those last two hints, since the are implications, rather than equivalences. So, I am now thinking that maybe it is ok to try and solve this problem with such implications (i.e. using inferencing) rather than using equivalences alone.

OK, so let's take that first hint, and let's try to change the goal to:

$a \Rightarrow \neg (b \lor c) \lor ((a \land b) \lor (a \land c))$

Because of Association, we can drop one pair of parentheses on the right side, and simply write this as :

$a \Rightarrow \neg (b \lor c) \lor (a \land b) \lor (a \land c)$

OK, now, I don't know if the following step is allowed by your professor, but using Contraposition, this is the same as:

$\neg (\neg (b \lor c) \lor (a \land b) \lor (a \land c)) \Rightarrow \neg a$

And by DeMorgan and possibly Double Negation (and again, ask your professor if you are allowed to use that), this is the same as:

$(b \lor c) \land \neg (a \land b) \land \neg (a \land c) \Rightarrow \neg a$

OK, so now comes the crucial step where I no longer use equivalences, but instead continue with inferencing:

Ok, so the last result shows that the goal has become to infer $\neg a$ from $(b \lor c) \land \neg (a \land b) \land \neg (a \land c)$. Ok, so let's assume that last statement, which means that we have:

1. $b \lor c$

2. $\neg (a \land b)$

3. $\neg (a \land c)$

Since by 1. we have either $b$ or $c$, let's consider each of those two possibilities:

If $b$, then we have (given 2): $b \land \neg (a \land b)$, which by Commutation (allowed by Professor?) is the same as $b \land \neg (b \land a)$ which by the last hint implies $\neg a$

If $c$, then using 3 we can similaely derive $\neg a$.

So, since either case leads to $\neg a$, we cannow defiitely say that $\neg a$, which was our goal.

OK, for now that is the best I can do. Again, I find your question very confusing since your professor did not spell out how to do this, and what rules you were able to use (I mean: if you can't assume something as simple as Distribution, then it is no longer clear what can be assumed!)