How to prove \$ A_d / A_{cm} = V_{cm} / \Delta V \$ in the CMRR equation
The second definition (\$CMRR=\frac{V_{cm}}{\Delta V}\$) must be wrong.
If \$\Delta V\$ is change in output caused by \$V_{cm}\$, then
$$\Delta V = A_{cm}\cdot V_{cm}\implies \frac{V_{cm}}{\Delta V}=\frac{1}{A_{cm}},$$
not \$CMRR\$.
\$CMRR\$ is telling you how much is \$V_d\$ amplified compared to how much \$V_{cm}\$ is amplified, not the amplification of \$V_{cm}\$ alone.
The way to measure it would be
$$CMRR = \frac{V_{cm}}{\Delta V} \cdot A_d$$
Which makes sense, since \$CMRR\$ should be a measure of &quot;how good the amplifier is at not amplifying common mode&quot;. Higher \$A_{cm}\$ shouldn't matter so much if \$A_d\$ is also very high, but if \$A_d\$ is low, then you also want very low \$A_{cm}\$ to get a good difference between wanted and unwanted signal.