Parsing this wff efficiently

Question

I have the wff:

$$\alpha = (((\neg(A_1\rightarrow (A_3\vee (\neg A_2))))\wedge(A_4\wedge(\neg A_1)))\rightarrow((\neg(A_3\vee A_2))\rightarrow(((\neg A_1)\wedge A_4)\vee A_3)))$$

I've already parsed through $12$ of the $16$ possibilities in the truth table. If we let

$$\beta = ((\neg(A_1\rightarrow (A_3\vee (\neg A_2))))\wedge(A_4\wedge(\neg A_1)))$$ and

$$\gamma = ((\neg(A_3\vee A_2))\rightarrow(((\neg A_1)\wedge A_4)\vee A_3))$$

Then $\alpha = (\beta \rightarrow \gamma)$.

I've already shown that $v(A_4)=F\rightarrow \bar{v}(\beta)=F\rightarrow \bar{v}(\alpha)=T$. I've also shown that $v(A_3)=T\rightarrow \bar{v}(\gamma)=T\rightarrow \bar{v}(\alpha)=T$. So $\alpha$ is true so far.

I just need to show what happens for the remaining $4$ cases when $v(A_3)=F$ and $v(A_4)=T$. I'm trying to do so efficiently, I know I can get the answer in a usual parsing algorithm. I noticed that when $A_3$ and $A_4$ are as such, it seems $\alpha$ is true for either $v(A_1)=T$ or $v(A_1)=F$ but I'm wondering if there's an efficient or clever way to realize that $\alpha$ is true no matter what.

Answer 1

$$\beta=\neg(A_1\to(A_3\lor\neg A_2))\land A_4\land\neg A_1$$ Since $\neg(A\to B)=A\land\neg B$ $$\beta=A_1\land\neg(A_3\lor\neg A_2)\land A_4\land\neg A_1=False$$ Since anything follows from False, $\alpha$ is a tautology.