Note: I'm aware this

**attempted**proof has issues, and am not claiming it is correct in any sense of the word.Consider the P=NP problem that asks whether every problem whose solution can be efficiently checked by a computer can also be efficiently solved by a computer.

P is not equal to NP because while a computer can efficiently (in polynomial time) check if sets P and NP are unequal, we cannot efficiently compute either P or NP.

Notes:

1. Computing P and NP would require determining whether every problem falls into the P or the NP class (itself a NP task).

2. It follows that class NP contains at least one more element (all others, plus the P=NP problem itself) than class P.

P is not equal to NP because while a computer can efficiently (in polynomial time) check if sets P and NP are unequal, we cannot efficiently compute either P or NP.

Notes:

1. Computing P and NP would require determining whether every problem falls into the P or the NP class (itself a NP task).

2. It follows that class NP contains at least one more element (all others, plus the P=NP problem itself) than class P.