0
$\begingroup$

I want to determine the logical converse of this result. I am confused.

The complex number $s=α+iβ$ is a solution of $f(s)=0$ and $α=1$ if and only if $g(s)≠0,h(s)=u(s)$ and $d(s)=v(s)$. Here the values of the mentioned functions are not important for the purpose of this function.

I know that if $a⇔b$, then converse of $a$⇔ converse of $b$

1 Answers 1

2

The converse is $f(s)\neq 0$ or $a\neq 1$ iff $g(s)=0$ or $h(s)\neq u(s)$ or $d(s)\neq v(s)$

This is because generally, $\neg \wedge\equiv \vee \neg$ where $\neg$ is the negation, $\wedge$ is "and" $\vee$ is "or"

Indeed $\neg (f(s)=0\wedge a=1)\equiv (\neg f(s) =0)\vee (\neg a=1)$ and $\neg (g(s)=0\wedge h(s)=u(s)\wedge d(s)=v(s))\equiv (\neg g(s) =0)\vee (\neg h(s)=u(s))\vee (\neg d(s)=v(s))$

It might be useful to know that $\neg \forall\equiv \exists\neg$ as well

  • 0
    Yes, and thank you. Things are now clear.2012-12-24