Mathematics: Can someone me with this truth table. i just don t understand how to do it.

Can someone me with this truth table. i just don

Ответ:

xoxzyanya61

05.01.2022

An implication $(p\implies q)$ is true if either the premise (p) is false, or both the premise and conclusion $(p\text{ and }q)$ are both true, and false otherwise.

$\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&&&&\\T&F&F&&&&\\F&T&T&&&&\\F&F&T&&&&\end{array}$

Negation $(\neg q)$ is straightforward; if a statement is true, then its negation is false, and vice versa.

$\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&&F\\T&F&F&T&&F\\F&T&T&F&&T\\F&F&T&T&&T\end{array}$

A conjunction $(p\land q)$ is true if both premises are true, and false otherwise.

$\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&F&F\\T&F&F&T&F&F\\F&T&T&F&F&T\\F&F&T&T&T&T\end{array}$

Finally, by the rules of implication, we can fill the last column:

$\begin{array}{c|c|c|c|c|c|c}p&q&p\implies q&\neg q&(p\implies q)\land\neg q&\neg p&(p\implies q)\land\neg q\implies\neg p\\T&T&T&F&F&F&T\\T&F&F&T&F&F&T\\F&T&T&F&F&T&T\\F&F&T&T&T&T&T\end{array}$

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Ответ:

blayneaafedt

11.06.2022

Mathematics

Bevel Marker: Single: Standard (recommended) 24x12x8: $625: $675: $725: $806: Economy : 24x12x6: $525: $575: $625: $655: Small : 20x10x6: $365: $405: $435: $465 pls mark branliest

Step-by-step explanation:

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Can someone me with this truth table. i just don't understand how to do it.

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