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.
Answer
- Minimal SOP
- F = AB + C
- Minimal POS
- F = (A + C)(B + C)
- Minterms
- F(A, B, C) = Σ(1, 3, 5, 6, 7)
- Maxterms
- F(A, B, C) = π(0, 2, 4)
Read as A'B'C + A'BC + AB'C + ABC' + ABC. It is a contingency (sometimes 1, sometimes 0).
Steps, law by law
F = A'B'C + A'BC + AB'C + ABC' + ABC
- = A'C(B' + B) + AB'C + ABC' + ABCDistributive law AB + AB' = A(B + B')
- = A'.C.1 + AB'C + ABC' + ABCComplement law A + A' = 1
- = A'C + AB'C + ABC' + ABCIdentity law A.1 = A
- = A'C + AC(B' + B) + ABC'Distributive law AB + AB' = A(B + B')
- = A'C + A.C.1 + ABC'Complement law A + A' = 1
- = A'C + AC + ABC'Identity law A.1 = A
- = C(A' + A) + ABC'Distributive law AB + AB' = A(B + B')
- = C.1 + ABC'Complement law A + A' = 1
- = C + ABC'Identity law A.1 = A
- = C + ABAbsorption law A + A'B = A + B
Convert to POS, step by step
Work out F' as a sum of products, then F = (F')' by De Morgan's law.
- F' = (C + AB)'Complement F' = (F)'
- F' = C'(AB)'De Morgan's law (A + B)' = A'B'
- F' = C'(A' + B')De Morgan's law (AB)' = A' + B'
- F' = C'A' + C'B'Distributive law A(B + C) = AB + AC
- F = (C'A' + C'B')'Complement again F = (F')'
- F = (C'A')'(C'B')'De Morgan's law (A + B)' = A'B'
- F = ((C')' + (A')')(C'B')'De Morgan's law (AB)' = A' + B'
- F = (C + (A')')(C'B')'Involution law (A')' = A
- F = (C + A)(C'B')'Involution law (A')' = A
- F = (C + A)((C')' + (B')')De Morgan's law (AB)' = A' + B'
- F = (C + A)(C + (B')')Involution law (A')' = A
- F = (C + A)(C + B)Involution law (A')' = A
Truth table
| A | B | C | AB'C | A'B'C + A'BC + AB'C | AB | C' | ABC' | A'B'C + A'BC + AB'C + ABC' | ABC | A'B'C + A'BC + AB'C + ABC' + ABC |
|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
K-map
- AB pair: cells 6, 7
- C quad: cells 1, 3, 5, 7
- (A + C) pair: cells 0, 2
- (B + C) pair: cells 0, 4
Canonical forms
SOP (sum of minterms)
F = A'B'C + A'BC + AB'C + ABC' + ABC
POS (product of maxterms)
F = (A + B + C)(A + B' + C)(A' + B + C)
Logic circuit





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