# Parsing this wff efficiently

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.

$$\beta=\neg(A_1\to(A_3\lor\neg A_2))\land A_4\land\neg A_1$$ 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.

• Wow, this is exactly what I’m looking for. Thank you Feb 21 at 1:34