If you're allowed to prove this the easy way (by showing you can use XOR and AND to create the set of AND, OR, and NOT), this is pretty straightforward.
x AND y = x AND y (of course) x OR y = (x XOR y) XOR (x AND y)
NOT x = x XOR 1
Also, (x AND y) XOR 1 is equivalent to x NAND y, which is a universal gate.
1 gate.
And, or, xor, xnor, nand, nor, not
Nand, nor, and, or, xor, nxor, not, true.
X-----Not--------------- | AND----------------| | |------------------ --------- |_ |__________ ---------OR-------OUTPUT | AND-----------------| Y-----|-Not-------------
XORing X with 1 gives X', i.e., NOT(X). If we are able to construct a NAND (AND) using XOR, we can also obtain AND (NAND) from it, which makes XOR a universal gate since inverted inputs to a NAND (AND) will give OR (NOR). However XOR is not a universal gate! Therefore we cannot obtain NAND (AND) using XOR. :-) By, Tirtha Sarathi Ghosh Class 10 IIT Kanpur Aspirant
No, XOR gate is a not a universal gate. There are basically two universal gates NAND and NOR.
Seven gates, they are: not, and, or, nor, nand, xor, xnor.
All other logic gates can be made using XOR and XNOR, but to get NOT, you need to do (input) XOR 1 or (input) XNOR 0, but with NAND, you don't need 1: (input) NAND (input).
AND OR NOT XOR
yes... xor is derived gate from primary gates
xor and xnor gates are derived from not gate
3*xor- two input
basic gates like XOR already exist in VHDL.
And, or, xor, xnor, nand, nor, not
1 gate.
Nand, nor, and, or, xor, nxor, not, true.
Here is your Answer... Basically NAND and NOR are the universal Gates, but the name or 8 Gates are given below: 1. AND 2. OR 3. NOR 4. NAND 5. XOR 6. XNOR 7. NOT 8. BUFFER Regards, M. Saad Shahid saadi.saad@gmail.com