NOR and NAND are far from the same thing.
AND is ab, but it is also (a' + b')'
NAND is (ab)', but it is also a' + b'
OR is a + b, but is also (a'b')'
NOR is (a + b)', but is also a'b'
Now your statement is that a'b'=(ab)'. Unfortunately, the inversion operator is not distributive across the unification of sets, this is easily shown with a Venn diagram.
I also remembered, you can get rid of a' by coding a NOR b and just making a and b read the same register. This operation would be no faster than a'. You can also put a' back into the instruction set, situated adjacent to the NOR function, and have a multiplexer at the NOR operator which either connects b or a to the second input. This only allows for a' to be found, though.
AND is ab, but it is also (a' + b')'
NAND is (ab)', but it is also a' + b'
OR is a + b, but is also (a'b')'
NOR is (a + b)', but is also a'b'
Now your statement is that a'b'=(ab)'. Unfortunately, the inversion operator is not distributive across the unification of sets, this is easily shown with a Venn diagram.
I also remembered, you can get rid of a' by coding a NOR b and just making a and b read the same register. This operation would be no faster than a'. You can also put a' back into the instruction set, situated adjacent to the NOR function, and have a multiplexer at the NOR operator which either connects b or a to the second input. This only allows for a' to be found, though.