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Following this answer by @DaftWullie, I give what I think of different cases of the CC-NOT gate:

  1. $I\otimes I\otimes I+ P_0\otimes P_0\otimes(U-I)$ is the CC-NOT operator when the first two controls are $0$.

  2. $I\otimes I\otimes I+ P_0\otimes P_1\otimes(U-I)$ is the CC-NOT operator when the first control is $0$ and second is $1$.

  3. $I\otimes I\otimes I+ P_1\otimes P_0\otimes(U-I)$ is the CC-NOT operator when the first control is $1$ and second is $0$.
  4. $I\otimes I\otimes I+ P_1\otimes P_1\otimes(U-I)$ is the CC-NOT operator when both controls are $1$.

What I understood from these cases is the following:

Consider the second case of $I\otimes I\otimes I+ P_0\otimes P_1\otimes(U-I)$. The first part of this operator does not check states of the first and second qubit i.e whether they are $00,01,10,11$ and operates the identity on the third qubit, which it did have to do when the states were $00,10,11$, but because of the Identity operated when the controls were $01$ is not what was intended so we add the extra part with projection operators $P_0$ and $P_1$ with the target qubit being acted by $(U-I)$ to negate the effect produced when we did not check the states and acted on the third qubit with the identity operator. I have similar arguments for the rest of the cases. Are they correct?

Here, $U$ is the unitary transform that I want to implement on the third register controlled by the first two, that is, the $X$ gate. $P_0$ and $P_1$ are the projection operators onto $|0\rangle$ and $|1\rangle$.

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Yes.

Both $I \otimes I \otimes I + P_0 \otimes P_1 \otimes (U-I)$ etc and the various CC-NOT are linear so in order to check equality of the operators, you only need to check on a basis. That is what you have done with the computational basis.

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