I have been stuck for a while trying to solve this, but it feels incomplete.

We first consider the distributivity and therefore start by proving that

$$a ∨ (b ∧ c) ╞ (a∨b)∧(a∨c)$$

Supose that for a value we have

$$a ∨ (b ∧ c) = True$$


$$a = True$$ or $$(b∧c) = True $$

What then?

Here I would make the jump to the right side of the equation, but I am unsure on how to concretely do it.

Also I am only proving this for one side of the equation, I will then have to do the oposite side and a similar proof for the other Distributive Law so a concrete answer would be nice.

  • $\begingroup$ Use the definition. The RHS is a conjuncition. What needs to happen for a conjunction to be true? Can you prove, in each case, that those things that need to happen, do happen indeed? $\endgroup$ – Git Gud Oct 6 '15 at 23:46
  • $\begingroup$ I am completely clueless as to what you mean, What even is RHS? $\endgroup$ – azthec Oct 6 '15 at 23:47
  • $\begingroup$ It's short for right-hand side (right-hand side of $\models$, i.e. $(a\lor b)\land (a\lor c)$). Is the term 'valuation' unfamiliar to you? $\endgroup$ – Git Gud Oct 6 '15 at 23:48
  • $\begingroup$ Many of these terms are unfamiliar to me, my country is not English speaking. I've figured out by conjunction you mean the And symbol, I cannot prove this with a truth table, as it would defeat the purpose. $\endgroup$ – azthec Oct 6 '15 at 23:51
  • $\begingroup$ But truth tables are part of the definition of $\models$. How are you defining the symbol $\models$? $\endgroup$ – Git Gud Oct 6 '15 at 23:54

You have disjoined the antecedent (the first statement) into two cases.   So use a proof by cases to show that both entail the desired consequent (the second statement).

$$\begin{align}a\vee(b\wedge c) & \vDash a, b\wedge c \\[2ex] a & \vDash a\vee b \\ a & \vDash a\vee c \\\hline a & \vDash (a\vee b)\wedge(a\vee c)\\[2ex]\vdots\\[2ex] a & \vDash (a\vee b)\wedge (a\vee c) \\ b\wedge c & \vDash (a\vee b)\wedge (a\vee c) \\\hline a\vee(b\wedge c) & \vDash (a\vee b)\wedge (a\vee c) \end{align}$$

For the first case, assume $a$ is true.   Then either $a$ or $b$ are true.   Also, either $a$ or $c$ are true.   So then both either $a$ or $b$ and either $a$ or $c$ hold true.   Hence the first case entails the desired result.

For the second case, assume both $b$ and $c$ are true.   Then ...

As both the first and second case entail the desired consequent, then the original antecedent entails the desired consequent.   And we are done.

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