PRASHNIKAप्रश्निका

Boolean algebra solver

Simplify a Boolean expression the way ISC Computer Science asks for it: the working law by law, the truth table, the K-map with its groups, the minimal SOP and POS forms, and the logic circuit - also with NAND or NOR gates only. Free, with nothing to sign up for.

Check that it equals another expression

Write NOT as A' (or ~A), AND as AB or A.B, OR as A + B; ⊕, => and <=> work too. Up to 6 variables. Or give minterms: F(A,B,C) = Σ(1,3,5) + d(7), or maxterms with π.

Try: A'B'C + A'BC + AB'C + ABC' + ABCF(A,B,C,D) = Σ(0,2,5,7,8,10,13,15)F(P,Q,R,S) = π(0,1,3,5,7,8,9,11) + d(2,13)(X + Z)(XY + YZ) + XZ + Y(A + B')(A' + B)A ⊕ B ⊕ C

Answer

Minimal SOP
F = AB + A'B'
Minimal POS
F = (A' + B)(A + B')
Minterms
F(A, B) = Σ(0, 3)
Maxterms
F(A, B) = π(1, 2)

Read as (A + B')(A' + B). It is a contingency (sometimes 1, sometimes 0).

Steps, law by law

F = (A + B')(A' + B)

  1. = A(A' + B) + B'(A' + B)Distributive law A(B + C) = AB + AC
  2. = AB + B'(A' + B)Absorption law A(A' + B) = AB
  3. = AB + B'A'Absorption law A(A' + B) = AB
Convert to POS, step by step

Work out F' as a sum of products, then F = (F')' by De Morgan's law.

  1. F' = (AB + B'A')'Complement F' = (F)'
  2. F' = (AB)'(B'A')'De Morgan's law (A + B)' = A'B'
  3. F' = (A' + B')(B'A')'De Morgan's law (AB)' = A' + B'
  4. F' = (A' + B')((B')' + (A')')De Morgan's law (AB)' = A' + B'
  5. F' = (A' + B')(B + (A')')Involution law (A')' = A
  6. F' = (A' + B')(B + A)Involution law (A')' = A
  7. F' = A'(B + A) + B'(B + A)Distributive law A(B + C) = AB + AC
  8. F' = A'B + B'(B + A)Absorption law A(A' + B) = AB
  9. F' = A'B + B'AAbsorption law A(A' + B) = AB
  10. F = (A'B + B'A)'Complement again F = (F')'
  11. F = (A'B)'(B'A)'De Morgan's law (A + B)' = A'B'
  12. F = ((A')' + B')(B'A)'De Morgan's law (AB)' = A' + B'
  13. F = (A + B')(B'A)'Involution law (A')' = A
  14. F = (A + B')((B')' + A')De Morgan's law (AB)' = A' + B'
  15. F = (A + B')(B + A')Involution law (A')' = A

Truth table

ABB'A + B'A'A' + B(A + B')(A' + B)
0011111
0100110
1011000
1101011

K-map

Grouping the 1s (SOP)
BA0101011231
  • AB single: cell 3
  • A'B' single: cell 0
Grouping the 0s (POS)
BA0101010203
  • (A' + B) single: cell 2
  • (A + B') single: cell 1

Canonical forms

SOP (sum of minterms)

F = A'B' + AB

POS (product of maxterms)

F = (A + B')(A' + B)

Logic circuit

Your expression as written Gate for gate as typed; A' is taken as an input line.
Logic circuit: Your expression as written
Minimal SOP AND, OR and NOT gates.
Logic circuit: Minimal SOP
Minimal POS OR, AND and NOT gates.
Logic circuit: Minimal POS
NAND gates only From the minimal SOP; a NOT is a NAND with its inputs joined.
Logic circuit: NAND gates only
NOR gates only From the minimal POS; a NOT is a NOR with its inputs joined.
Logic circuit: NOR gates only

Practise on real ISC questions