Solve truth table task

Question

I am solving next task: $$\left(x \vee y\right) \rightarrow \bar{x}$$

I assume that line above $\bar{x}$ makes it opposite of $x$ and my solving is this: $$\begin{array}{|cc|c|c|c|} \hline x &y & x\lor y &x & \left(x \vee y\right) \rightarrow \bar{x} \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{F} \\ \text{T} & \text{F} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \end{array}$$ Does my calculations correct? If not why?

EDITED

Answer 1

$$\begin{align} &x &y &&x\lor y &&\bar x &&(x \vee y) \to\bar x \\ &T &T &&T &&\color{red}F &&F \\ &T &F &&T &&F &&F\\ &F &T &&T &&T &&T\\ &F &F &&F &&T &&T \end{align}$$

$(x \vee y)$ is your hypothesis, the conditional statement is only false when the hypothesis $(x \vee y)$ is true, and your conclusion ($\bar x$) is false. When you hypothesis is false, the conditional statement is true by default.

Answer 2

$$\begin{align} &x &y &&x\lor y &&\bar{x} &&(x \vee y) \rightarrow \bar{x} \\ &T &T &&T &&F &&F \\ &T &F &&T &&F &&F\\ &F &T &&T &&T &&T\\ &F &F &&F &&T &&T \end{align}$$

Think of it this way: If the premise is true, then the conclusion must be true as well. The premise being true is necessary for the conclusion to be true. For your argument to be valid, if the premise is true then the conclusion cannot be false.