Verifying Truth Tables

Question

would someone please verify the following truth table? If there is something innacurate, would someone please explain why?

Thank You! $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \color{black}{L} & \color{black}{G} & \color{black}{H} & \lnot H & (G\lor\lnot H)& \lnot(G\land L) &(G\lor \lnot H )\land \lnot(G\land L) \\ \hline \color{}{T} & \color{}{T} & \color{}{T} & \color{}{F} & \color{}{T}& \color{}{F}&\color{}{F} \\ \hline \color{black}{T} & \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{F}&\color{black}{F} \\ \hline \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{F}& \color{black}{T} &\color{black}{F} \\ \hline \color{black}{T} & \color{black}{F} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{T} & \color{black}{T} & \color{black}{F} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{F}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{F} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \end{array}$$

Answer 1

Since you provided a nicely formatted table, I am happy to provide my own solutions. I marked any mistake, and entries that - while being correct - were affected by previous mistakes, in red and hopefully caught all of them (;

\begin{array}{|c|c|c|c|c|c|c|c|}\hline \color{black}{L} & \color{black}{G} & \color{black}{H} & \lnot H & (G\lor\lnot H)& \lnot(G\land L) &(G\lor \lnot H )\land \lnot(G\land L) \\ \hline \color{}{T} & \color{}{T} & \color{}{T} & \color{}{F} & \color{}{T}& \color{}{F}&\color{}{F} \\ \hline \color{black}{T} & \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{F}&\color{black}{F} \\ \hline \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{F}& \color{black}{T} &\color{black}{F} \\ \hline \color{black}{T} & \color{black}{F} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{T} & \color{black}{T} & \color{black}{F} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{T} & \color{black}{T}& \color{black}{T}&\color{black}{T} \\ \hline \color{black}{F} & \color{black}{F} & \color{black}{T} & \color{black}{F} & \color{black}{F}& \color{black}{T}&\color{red}{F} \\ \hline \color{black}{F} & \color{black}{F} & \color{black}{F} & \color{red}{T} & \color{red}{T}& \color{black}{T}&\color{red}{T} \\ \hline \end{array}

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edit: Since you also asked for an explanation, let us enumerate the mistakes from top to bottom, left to right:

1. A conjunction $\phi \wedge \psi$ is true iff both $\phi$ and $\psi$ are true. Hence, since $(G \vee \neg H)$ is false $(G \vee \neg H) \wedge \neg(G \wedge L)$ must also be false.
2. Since $H$ is false, $\neg H$ is true.
3. Your post already, correctly, declared $(G \vee \neg H)$ to be true. However, since you accidentally declared $\neg H$ to be false, this wasn't justified.
4. I only marked this as a mistake, because it depended on item 3. which in turn depended on 2.