Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — simplifying with absorption
Simplify A+AB and say how many gates are saved.
- 1Take out the common factor A.The first move in almost every simplification.
- 2A+AB=A(1+B)
- 31+B=1The null law — an OR with a guaranteed 1 is always 1.
- 4A⋅1=AIdentity law.
- 5So A+AB=A: the B input is irrelevant.An AND gate and an OR gate are both removed — two gates saved.
Example 2 — using the second distributive law
Simplify A+A′B.
- 1Apply A+BC=(A+B)(A+C) with B→A′ and C→B.This law has no counterpart in ordinary algebra.
- 2A+A′B=(A+A′)(A+B)
- 3A+A′=1Complement law.
- 41⋅(A+B)=A+B
- 5Check the truth table: both give 0,1,1,1 ✓A NOT gate and an AND gate are eliminated.
Example 3 — verifying De Morgan's law
Verify that (A⋅B)′=A′+B′ using a truth table.
- 1A=0,B=0: AB=0, so (AB)′=1; and A′+B′=1+1=1 ✓
- 2A=0,B=1: AB=0, so (AB)′=1; and 1+0=1 ✓
- 3A=1,B=0: AB=0, so (AB)′=1; and 0+1=1 ✓
- 4A=1,B=1: AB=1, so (AB)′=0; and 0+0=0 ✓
- 5The columns agree in all four rows, so the law holds.Note A′⋅B′ would give 1,0,0,0 — quite different.