Level 4: At least one

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Task

Goal: build a logical OR gate.

OR outputs 1 if at least one input is 1. It outputs 0 only when both inputs are 0.

The main idea: OR is built from what you already have. Invert both inputs, combine them with AND, then invert again. This is De Morgan's law: at least one is on means it is not true that both are off.

Path to this level

The competencies this level develops and their formation levels.

NAND universality · ур. 2, 3 Boolean composition · ур. 3

Related Materials

  • OR — logical addition

Solution for self-check

Try building the circuit yourself first. The solution below is a way to check yourself if you're stuck or have already solved it.

What you should get: the output is 1 when at least one input is 1.

Formula: A OR B = NOT(NOT A AND NOT B). In words: at least one is on means it is not true that both are off.

  1. Place two NOT gates and feed A and B into them separately.
  2. Feed both NOT outputs into the AND inputs.
  3. Pass the AND output through a third NOT.
  4. Connect the third NOT output to Output.

Why it works: AND outputs 1 only when both inputs are 0, which means A and B are both 0. The third NOT flips that result, so 1 appears in all other cases.

Solution circuit for level 4 — At least one
The authored solution for level 4 in the simulator (dark theme)

Check yourself

What happens if you remove the third NOT?

You get NOR: 1 only when both inputs are 0, and 0 whenever at least one input is 1.