Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) %3D = (x + x')(y + x')(x + z)(y + z) = (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore,
Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) %3D = (x + x')(y + x')(x + z)(y + z) = (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore,
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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
Transcribed Image Text:EXAMPLE 2.5
Express the Boolean function F = xy + x'z as a product of maxterms. First, convert
the function into OR terms by using the distributive law:
F = xy + x'z = (xy + x')(xy + z)
= (x + x')(y + x')(x + z)(y + z)
(x' + y)(x + z)(y + z)
The function has three variables: x, y, and z. Each OR term is missing one variable;
therefore,
x' + y = x' + y + zz'
(x' + y + z)(x' + y + z')
x + z = x + z + yy' = (x + y + z)(x + y' + z)
%3D
y + z = y + z + xx'
(x + y + z)(x' + y + z)
%3D
Combining all the terms and removing those which appear more than once, we finally
obtain
F — (х + у z)(x + y' +z)(x' + у + z)(x' + у + z')
= M,M2M4M5
F(x, y, z) = II(0, 2, 4, 5)
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