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Converse nonimplication

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

Dual

$::{\begin{aligned}dual(q\nleftarrow p)&=dual(q'p)\\&=q'+p\\&=((q'+p)')'\\&=(qp')'\\\end{aligned}}$

Table summarizing the main boolean laws

S.No	Law	Expression
1	Identity	${\begin{aligned}x\lor 0&=x\\x\land 1&=x\end{aligned}}$
2	Annihilator	${\begin{aligned}x\lor 1&=1\\x\land 0&=0\end{aligned}}$
3	Idempotence	${\begin{aligned}x\lor x&=x\\x\land x&=x\end{aligned}}$
4	Involution	${\begin{aligned}\lnot \lnot x&=x\end{aligned}}$
5	Complementarity	${\begin{aligned}x\lor \lnot x&=1\\x\land \lnot x&=0\end{aligned}}$
6	Commutative	${\begin{aligned}x\lor y&=y\lor x\\x\land y&=y\land x\end{aligned}}$
7	Associative	${\begin{aligned}(x\lor y)\lor z&=x\lor (y\lor z)\\(x\land y)\land z&=x\land (y\land z)\end{aligned}}$
8	Distributive	${\begin{aligned}x\land (y\lor z)&=(x\land y)\lor (x\land z)\\x\lor (y\land z)&=(x\lor y)\land (x\lor z)\end{aligned}}$
9	Absorption	${\begin{aligned}x\lor (x\land y)&=x\\x\land (x\lor y)&=x\end{aligned}}$
10	De Morgan	${\begin{aligned}\lnot (x\lor y)&=(\lnot x)\land (\lnot y)\\\lnot (x\land y)&=(\lnot x)\lor (\lnot y)\end{aligned}}$

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:

${\begin{aligned}I_{1}{\dot {\omega }}_{1}&=(I_{2}-I_{3})\omega _{2}\omega _{3}~~~~~~~~~~~~~~~~~~~~{\text{(1)}}\\I_{2}{\dot {\omega }}_{2}&=(I_{3}-I_{1})\omega _{3}\omega _{1}~~~~~~~~~~~~~~~~~~~~{\text{(2)}}\\I_{3}{\dot {\omega }}_{3}&=(I_{1}-I_{2})\omega _{1}\omega _{2}~~~~~~~~~~~~~~~~~~~~{\text{(3)}}\end{aligned}}$

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),

${\begin{aligned}I_{2}I_{3}{\ddot {\omega }}_{2}&=(I_{3}-I_{1})(I_{1}-I_{2})\omega _{1}\omega _{2}\\{\text{i.e.}}~~~~{\ddot {\omega }}_{2}&={\text{(negative quantity)}}\times \omega _{2}\end{aligned}}$

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),

${\begin{aligned}I_{1}I_{3}{\ddot {\omega }}_{1}&=(I_{2}-I_{3})(I_{1}-I_{2})\omega _{1}\omega _{2}\\{\text{i.e.}}~~~~{\ddot {\omega }}_{1}&={\text{(positive quantity)}}\times \omega _{1}\end{aligned}}$

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'.