උත්තරේ හරිද

Commex

Well-known member
  • Nov 29, 2019
    2,020
    1,588
    113
    5BE2FDCD-1D9D-490F-9222-412BBAA08C7F.jpeg

    මේකට අදාලව කානෝ මැප් එකක් ඇදලා සුලු කරාම එන්නේ මේ උත්තරේද බන්. අද එක්සෑම් තිබුනා.මේක මටආවේ. හරියටම ශුවර් නැති නිසා ඇහුවෙ :)
     
    • Like
    Reactions: Gwynbleidd

    imhotep

    Well-known member
  • Mar 29, 2017
    14,856
    8
    35,444
    113
    Hope this helps.....

    You need to understand the Boolean algebra properties....

    Commutative AB=BA , A+B=B+A
    Associative ABC=(AB)C=A(BC) , A+B+C=(A+B)+C=A+(B+C)
    Distributive A(B+C)=AB+AC

    Some basic theorems.....

    A · 0 = 0 , A · 1=A
    A+0 = A , A+1=1
    A · A=A , A · A̅ =0
    A+A=A , A+ A̅ =1

    (A̅ )' = A (A bar bar)

    A+AB = A
    A+ A̅ B = A+B
    (A+B)(A+C) = A+BC
    (AB)' = A̅ + B̅ , AB bar
    (A+B)' = A̅ · B̅ , (A+B) bar

    (For ease of typing ' used as bar over the bracketed symbols)

    Apply these rules to simplify any expression.
     
    Last edited:

    Commex

    Well-known member
  • Nov 29, 2019
    2,020
    1,588
    113
    Ummmmaaa :love:
    1598520117871.png

    Hope this helps.....

    You need to understand the Boolean algebra properties....

    Commutative AB=BA , A+B=B+A
    Associative ABC=(AB)C=A(BC) , A+B+C=(A+B)+C=A+(B+C)
    Distributive A(B+C)=AB+AC

    Some basic theorems.....

    A · 0 = 0 , A · 1=A
    A+0+A , A+1=1
    A · A=A , A. · A̅ =0
    A+A=A , A+ A̅ =1

    (A̅ )' = A (A bar bar)

    A+AB = A
    A+ A̅ B = A+B
    (A+B)(A+C) = A+BC
    (AB)' = A̅ + B̅ , AB bar
    (A+B)' = A̅ · B̅ , (A+B) bar

    (For ease of typing ' used as bar over the bracketed symbols)

    Apply these rules to simplify any expression.
    බොහොම ස්තූතියි මචං ♥️
    1598520250401.png

    1598520268708.png
     

    imhotep

    Well-known member
  • Mar 29, 2017
    14,856
    8
    35,444
    113
    Without a K-map, the simplification.....

    Z = A'B + BC' + BC + AB'C'
    = A'B + (BC'+BC') + BC + AB'C' (because BC'=BC'+BC')
    = A'B + (BC'+BC) + (BC'+AB'C')
    = A'B + B(C+C') + C'(B+AB')
    = A'B + B + C'(B+A) (because B+AB' = B+A)
    = B(A'+1) + C'(B+A) (because A'+1 = 1)
    = B + C'(B+A)
    = B + BC' + AC'
    = B(1 + C') + AC' (because 1+C' = 1)
    = B +AC'