Natural deduction proof: $A \vee (B \wedge C) ├ (A \vee B) \wedge (A \vee C)$

Question

I assume that I need to set a hypothesis somewhere in the process, but I don't know how.

1   A V (B Ʌ C)     Premiss
2   B Ʌ C           1 (VE)
3   B               2 (ɅE)
4   C               2 (ɅE)
5   A V B           3 (VI)

The above is the formulas that I have written so far, but I'm clearly stuck and not sure where / how to continue. (I also doubt if line 5 ($A \vee B$) is the right one)

How should I tackle this problem?

Answer 1

1. You got the $\lor E$ rule wrong. From $A \lor (B \land C)$ you can not directly infer $A$ and $(B \land C)$ - that would be $\land E$. Instead, you need to open two subproofs where you assume $A$ and $B \land C$ respectively, then when both subproofs yield the same conclusion $(A \lor B) \land (A \lor C)$, this will be the conclusion of $\lor E$, where the assumptions $A$ and $B \land C$ can be discarded. The idea is "If we know that either $A$ or $B \land C$ hold, and from both scenarios proposition $(A \lor B) \land (A \lor C)$ would follow, then we can conclude $(A \lor B) \land (A \lor C)$."
2. To continue the proof, you proceed in the same way for $C$ as you did for $B$.

Full proof:

Answer 2

1   A V (B Ʌ C)     Premiss
2   B Ʌ C           1 (VE)
3   B               2 (ɅE)
4   C               2 (ɅE)
5   A V B           3 (VI)
6   A V C           4 (VI)
7   (A V B) Ʌ (A V C)  5,6
8   A               1 (VE)
9   A V B           8 (VI)
10  A V C           8 (VI)
11  (A V B) Ʌ (A V C)  9,10

I'm not correct on the formalism but you were just about there.

Answer 3

Apparently you wanted to eliminate A from A v ( B&C) , but you can do this only if you first have ~A. Since ~ A is not given as premise, the best way is to introduce ~A as hypothesis, and then to use the disjunctive syllogism rule inside a conditional proof.

Here by " disjunctive syllogism" I mean the rule

" from X v Y and ~X, infer : Y "

Answer 4

$$\frac{\frac{A\vdash A\lor B \; \; A\vdash A\lor C}{A\vdash (A\lor B)\land (A\lor C)} \;\;\frac{\frac{B\land C\vdash B \;\; B\vdash A\lor B}{B\land C\vdash A\lor B} \frac{B\land C\vdash C \;\; C\vdash A\lor C}{B\land C\vdash A\lor C}}{B\land C\vdash (A\lor B)\land (A\lor C)}}{A\lor(B\land C)\vdash (A\lor B)\land (A\lor C)}$$

This is a proof using sequent calculus, which may not be what you're looking for; but all proof systems at this level are essentially the same, and so if you can identify the rules that I'm using, and relate them to the rules of the system you want to use, then it can help you nonetheless