User:Roshan220195/sandbox

Converse nonimplication
$$ \begin{align} (r \nleftarrow q) \nleftarrow p &= r'q \nleftarrow p \qquad \qquad \qquad ~ \text{(by definition)} \\ &= (r'q)'p \qquad \qquad \qquad \text{(by definition)} \\ &= (r + q')p \qquad \qquad \text{(De Morgan's laws)} \\ &= (r + r'q')p \qquad \qquad ~ \text{(Absorption law)} \\ &= rp + r'q'p \\ &= rp + r'(q \nleftarrow p) \qquad \text{(by definition)} \\ &= rp + r \nleftarrow (q \nleftarrow p) \qquad ~ \text{(by definition)} \\ \end{align} $$

Dual
$$
 * \begin{align}

dual(q \nleftarrow p) &= dual(q'p) \\ &= q' + p \\ &= ((q' + p)')' \\ &= (qp')' \\

\end{align} $$

Qualitative Proof
The tennis racket theorem can be qualitatively analysed with the help of Euler's equations.

Under torque free conditions, they take the following form:

\begin{align} I_1\dot{\omega}_{1}&=(I_2-I_3)\omega_2\omega_3\text{(1)}\\ I_2\dot{\omega}_{2}&=(I_3-I_1)\omega_3\omega_1\text{(2)}\\ I_3\dot{\omega}_{3}&=(I_1-I_2)\omega_1\omega_2\text{(3)} \end{align} $$

Let $$ I_1 > I_2 > I_3 $$

Consider the situation when the object is rotating about axis with moment of inertia $$I_1$$. To determine the nature of equilibrium, assume small initial angular velocities along the other two axes. As a result, according to equation (1), $$~\dot{\omega}_{1}$$ is very small. Therefore the time dependence of $$~\omega_1$$ may be neglected.

Now, differentiating equation (2) and substituting $$\dot{\omega}_3$$ from equation (3),

\begin{align} I_2 I_3 \ddot{\omega}_{2}&= (I_3-I_1) (I_1-I_2) \omega_1\omega_{2}\\ \text{i.e.}~ \ddot{\omega}_2 &= \text{(negative quantity)} \times \omega_2 \end{align} $$

Note that $$\omega_2$$ is being opposed and so rotation around this axis is stable for the object.

Similar reasoning also gives that rotation around axis with moment of inertia $$I_3$$ is also stable.

Now apply the same thing to axis with moment of inertia $$I_2$$. This time $$\dot{\omega}_{2}$$ is very small. Therefore the time dependence of $$~\omega_2$$ may be neglected.

Now, differentiating equation (1) and substituting $$\dot{\omega}_3$$ from equation (3),

\begin{align} I_1 I_3 \ddot{\omega}_{1}&= (I_2-I_3) (I_1-I_2) \omega_1\omega_{2}\\ \text{i.e.}~ \ddot{\omega}_1 &= \text{(positive quantity)} \times \omega_1 \end{align} $$

Note that $$\omega_1$$ is not opposed and so rotation around this axis is unstable. Therefore even a small disturbance along other axes causes the object to 'flip'.