Q. DeMorgan’s first theorem shows the equivalence of (Solved)
1. or gate and exclusive or gate.
2. nor gate and bubbled and gate.
3. nor gate and nand gate.
4. nandgate and not gate
- b. nor gate and bubbled and gate.
1. or gate and exclusive or gate.
2. nor gate and bubbled and gate.
3. nor gate and nand gate.
4. nandgate and not gate