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

• Do you mean the claim $$(\bar{x} \vee y) \rightarrow x$$ ? Please note, that you can use \bar{x} in LATEX to generate \bar{x}. – 3nondatur Feb 4 '18 at 10:30 • @3nondatur I will EDIT – IntoTheDeep Feb 4 '18 at 10:31 ## 2 Answers \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.

• Can you please check edited answer – IntoTheDeep Feb 4 '18 at 10:33
• Noticed you have edited the question. Have edited my answer and provided an explanation for you as well. :) – Icycarus Feb 4 '18 at 10:37

\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.

• Why this answer is different that @lcycarus's – IntoTheDeep Feb 4 '18 at 10:39
• I've noticed your edit of the the fourth column. I've edited and our answers match. – OGC Feb 4 '18 at 10:42