1 Kingdom of Saudi Arabia
Ministry of Education Umm AlQura University
Adam University College, female branch Computer Science Department
Ψ©ΩΨ―ΩΨΉΨ³ΩΨ§ Ψ©ΩΨ¨Ψ±ΨΉΩΨ§ Ψ©ΩΩΩ Ω ΩΨ§ Ω ΩΩΨΉΨͺΩΨ§ Ψ©Ψ±Ψ§Ψ²Ω Ψ©ΨΉΩ Ψ§Ψ¬ ΩΨ±ΩΩΨ§ Ω Ψ£ ΨͺΨ§Ψ¨ΩΨ§Ψ·ΩΨ§ Ψ±Ψ·Ψ΄ ΨΩ ΨΆΨ£Ψ¨ Ψ©ΩΨΉΩ Ψ§Ψ¬ΩΨ§ Ψ©ΩΩΩΩΨ§
ΩΩΩΨ’Ψ§ Ψ¨Ψ³Ψ§ΨΩΨ§ Ω ΩΩΨΉ Ω Ψ³Ω
First Semester of 2017-2018 Academic Year
Logic Analysis and Design Course, 6803213-3 Second DeMorgansβ Law Proof
The Theorem:
π΄ β π΅
Μ Μ Μ Μ Μ Μ Μ = π΄Μ + π΅Μ
The Answer:
For any theorm π = π, if we can show that πΜ π = 0, and that πΜ + π = 1 then by the complement identities, π΄Μ π΄ = 0 and π΄Μ + π΄ = 1, πΜ = πΜ .
By the uniqueness of the complement, π = π.
Thus the proof consists of showing that (π΄ β π΅) β (π΄Μ + π΅Μ ) = 0; and also that (π΄ β π΅) + (π΄Μ + π΅Μ ) = 1.
Prove: (π΄ β π΅) β (π΄Μ + π΅Μ ) = 0
(π΄ β π΅) β (π΄Μ + π΅Μ ) = (π΄ β π΅) β π΄Μ + (π΄ β π΅) β π΅Μ By Distributive identitiy = (π΄ β π΄Μ ) β π΅ + π΄ β (π΅ β π΅Μ ) By associativitiy identitiy = 0 β π΅ + π΄ β 0 By complement identity = 0 + 0 By nulity theorem = 0 By identity theorem (π΄ β π΅) β (π΄Μ + π΅Μ ) = 0
Prove: (π΄ β π΅) + (π΄Μ + π΅Μ ) = 1
(π΄ β π΅) + (π΄Μ + π΅Μ ) = (π΄ + π΄Μ + π΅Μ ) β (π΅ + π΄Μ + π΅Μ ) By Distributive identitiy (π΄ β π΅) + (π΄Μ + π΅Μ ) = (π΄ + π΄Μ + π΅Μ ) β (π΅ + π΅Μ + π΄Μ ) By associativitiy identitiy = (1 + π΅Μ ) β (1 + π΄Μ ) By complement identity = 1 β 1 By nulity theorem = 1 By identity theorem (π΄ β π΅) + (π΄Μ + π΅Μ ) = 1
Since (π¨ β π©) β (π¨Μ + π©Μ ) = π, and (π¨ β π©) + (π¨Μ + π©Μ ) = π, π¨ β π© is complement of π¨Μ + π©Μ meaning that π¨ β π© = (π¨Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ + π©Μ ); Thus π¨ β π©Μ Μ Μ Μ Μ Μ Μ = (π¨Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ + π©Μ )
.
The involution theorm states that π¨Μ Μ = π¨. Thus by the involution theorm, (π¨Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ + π©Μ ) = π¨Μ + π©Μ . This proves the above DeMorganβs Theorem.
2 References:
[1] http://www2.nau.edu/~sh295/EE110/deMorganproof.html
Good Luck my Great Students π T.Mariah Sami Ahmed Khayat
Teacher Assistant @ Adam University College [email protected]