...
ʶ˨ˠˢ˟ˮ˞%RROH
µµ (George Boole, 1815-1864)
1847 µ µ µ µ (
" µ"). µ µµ !# # # . #µ # µ , # , # ! $µ# ! µ % %µ% % , # # !µ #%
&% µ.
x
To 1854 o Boole & Boole
x
To 1938 o Shannon # x
' Boole µ # µ# µ % !% * µ & + <
% &µ Huntington1:
1. +# % +
+# % <
2. 0& % +: (x+0 = 0+x = x)
0& % <
µ µ 1: (x < 1 = 1 < x = x)
3. µ# % + : x + y = y + x
µ# % <:
x
< y=y < x
< µ# % & +. <#:
x < (y+z) = (x < y) + (x < z)
' & + µ# % < . <#:
4. ' &
1
> µµ (1874-1952) &µ 1933
5
...
x + y < z = (x + y)
< (x + z)
5. ! x
µ#%µ :
B
!
xc  B
µ
x x c 1, x < x c 0
6. * ! ! µ !.
<# ! ! x, y B x ? y
µ 1
µ 2
˂˲˞ˡ˦˧˛ʶ˨ˠˢ˟ˮ˞%RROH
µ
@% * = {0, 1} % & “+” “ < ” % &#:
x
y
x+y
x <y
1
1
1
1
1
0
1
0
0
1
1
0
0
0
0
0
µ# (B, ,<) &µ Huntington.
µ * & µ#%µ % &#:
' # µ# (B, ,<) ! µ# & µ#%µ # Boole.
µ 5
6
PÐnakac idiot twn thc duadik c ˆlgebrac Boole
Gia thn duadik ˆlgebra Boole, isqÔoun ta parakˆtw:
¾
x+y =y+x
(antimetajetikìthta)
x·y =y·x
¾
x + (y + z) = (x + y) + z
(prosetairistikìthta)
x · (y · z) = ¾
(x · y) · z
x+x=x
(adunamÐa)
x·x=x
¾
x + (x · y) = x
(aporrofhtikìthta)
x · (x ∨ y) = x
kai epiplèon
¾
x · (y + z) = x · y + x · z
(epimeristikìthta)
x + y · z = (x + y) · (x + z)
IsqÔoun epÐshc oi parakˆtw idiìthtec:
i. (x0 )0 = x
ii. x + 0 = x kai x + 1 = 1.
iii. x · 0 = 0 kai x · 1 = x.
iv. x + x0 = 1 kai x · x0 = 0.
v. x + x = x kai x · x = x.
vi. x + x0 · y = x + y . ¾
(x + y)0 = x0 · y 0
vii.
(tÔpoi De Morgan).
(x · y)0 = x0 + y 0
Oi apodeÐxeic twn idiot twn gÐnontai me pÐnakec me qr sh twn axiwmˆtwn tou Huntington.
ParadeÐgmata
1. Apìdeixh thc prosetairistik c idiìthtac
x + (y + z) = (x + y) + z
x
1
1
1
1
0
0
0
0
y
1
1
0
0
1
1
0
0
z
1
0
1
0
1
0
1
0
y+z
1
1
1
0
1
1
1
0
x + (y + z)
1
1
1
1
1
1
1
0
x+y
1
1
1
1
1
1
0
0
2. Apìdeixh twn epimeristik¸n idiot twn
x · (y + z) = x · y + x · z
7
(x + y) + z
1
1
1
1
1
1
1
0
x
1
1
1
1
0
0
0
0
y
1
1
0
0
1
1
0
0
z
1
0
1
0
1
0
1
0
y+z
1
1
1
0
1
1
1
0
x(y + z) xy
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
xz
1
0
1
0
0
0
0
0
xy + xz
1
1
1
0
0
0
0
0
x+y · z = (x+y) · (x+z)
x
1
1
1
1
0
0
0
0
y
1
1
0
0
1
1
0
0
z
1
0
1
0
1
0
1
0
yz
1
0
0
0
1
0
0
0
x + yz
1
1
1
1
1
0
0
0
x+y
1
1
1
1
1
1
0
0
x+z
1
1
1
1
1
0
1
0
3. Apìdeixh twn idiot twn iv
x + x0 = 1 kai x · x0 = 0
x
0
1
x0
1
0
x + x0
1
1
xx0
0
0
4. Apìdeixh thc aporrofhtik c idiìthtac
x+x·y =x:
x + xy = x1 + xy = x(1 + y) = x1 = x.
5. Apìdeixh thc idiìthtac vi.
x0 + x · y = x0 + y :
x0 + xy = (x0 + x)(x0 + y) = 1(x0 + y) = x0 + y
Parˆdeigma 1 Na lujeÐ h exÐswsh Boole
x0 y + xy 0 = 0.
8
(x + y)(y + z)
1
1
1
1
1
0
0
0
x
1
1
0
0
y
1
0
1
0
x0
0
0
1
1
x0 y
0
0
1
0
y0
0
1
0
1
xy 0
0
1
0
0
x0 y + xy 0
0
1
1
0


x = y = 1
'Ara 

x = y = 0.
Parˆdeigma 2 Na lujeÐ h exÐswsh Boole
xz 0 + x0 yz + y 0 z 0 = 1.
'Estw F = xz 0 + x0 yz + y 0 z 0 .
x
1
1
1
1
0
0
0
0
y
1
1
0
0
1
1
0
0
z
1
0
1
0
1
0
1
0
z0
0
1
0
1
0
1
0
1
xz 0
0
1
0
1
0
0
0
0
x0
0
0
0
0
1
1
1
1
yz
1
0
0
0
1
0
0
0
x0 yz
0
0
0
0
1
0
0
0
y0
0
0
1
1
0
0
1
1
y0z0
0
0
0
1
0
0
0
1
xz 0 + x0 yz
0
1
0
1
1
0
0
0


x = y = 1, z = 0










x = 1, y = z = 0
'Ara, 


x = 0, y = z = 1








x = y = z = 0.
Parˆdeigma 3 Na lujeÐ (kai diereunhjeÐ) h exÐswsh Boole
ax + bx0 = 0, ìpou a, b ∈ B.
a
1
1
0
0
b
1
0
1
0
ax + bx0
x + x0 (=1)
x
x0
0
→ 'Ara adÔnath.
→ 'Ara x = 0.
→ 'Ara x0 = 0, (dhlad x = 1).
→ 'Ara tautìthta, dhlad isqÔei
gia kˆje x, (dhlad isqÔei
gia x = 0 kai gia x = 1).
9
F
0
1
0
1
1
0
0
1
Sust mata
½
Na lujeÐ to sÔsthma Boole
x
1
1
0
0
y
1
0
1
0
x0
0
0
1
1
x0 + xy 0
x + xy
y0
0
1
0
1
xy 0
0
1
0
0
¾
=1
=0
xy
1
0
0
0
.
x0 + xy 0
0
1
1
1
x + xy
1
1
0
0
'Ara (x, y) = (0, 1) (x, y) = (0, 0).
Sunart seic Boole
Kˆje sunˆrthsh f : B n → B lègetai
Parˆdeigma 1
sunˆrthsh Boole.
f : B 2 → B me f (x, y) = xy 0 + x0 y.
H f paÐrnei tic timèc:
f (1, 1) = 1 · 0 + 0 · 1 = 0 + 0 = 0,
f (1, 0) = 1 · 1 + 0 · 0 = 1 + 0 = 1,
f (0, 1) = 0 · 0 + 1 · 1 = 0 + 1 = 1,
f (0, 0) = 0 · 1 + 1 · 0 = 0 + 0 = 0,
(me pÐnaka)
x
1
1
0
0
y
1
0
1
0
y0
0
1
0
1
xy 0
0
1
0
0
x0
0
0
1
1
Parˆdeigma 2
x0 y
0
0
1
0
f
0
1
1
0
f : B 3 → B me f (x, y, z) = xy + z 0 .
H f paÐrnei tic timèc:
f (1, 1, 1) = 1 · 1 + 0 = 1 + 0 = 1.
f (1, 1, 0) = 1 · 1 + 1 = 1 + 1 = 1.
f (1, 0, 1) = 1 · 0 + 0 = 0 + 0 = 0. k.lp.
Efarmogèc
1. Diakìptec
Diakìptec parˆllhloi −→ +
1+1=1
1+0=1
(0 + 1 = 1)
10
0+0=0
Diakìptec se seirˆ−→ ·
1·1=1
2. DÐpola
1·0=0
(0 · 1 = 0)
0·0=0
Se kˆje sunˆrthsh Boole antistoiqeÐ èna dÐpolo kai antÐstrofa:
Sto dÐpolo
x’
y’
x
z
y
antistoiqeÐ h sunˆrthsh
f (x, y, z) = x0 y 0 + z(x + y).
Sto dÐpolo
antistoiqeÐ h sunˆrthsh
f (x, y, z, w) = xy + xwz + y 0 wy + y 0 z
= xy + xwz + y 0 yw + y 0 z
= xy + xwz + 0w + y 0 z
= xy + xwz + y 0 z.
AntÐstrofa:
Sth sunˆrthsh
f (x, y, z) = xy 0 z 0 + x0 yz + xyz
antistoiqeÐ to dÐpolo
Allˆ
f (x, y, z) = xy 0 z 0 + (x0 + x)yz
= xy 0 z 0 + 1 · yz
= xy 0 z 0 + yz,
opìte paÐrnoume to antÐstoiqo (aploÔstero) dÐpolo
11
x
y
w
y’
z
x
y’
z’
x’
y
z
x
y
z
Ask seic
1) Me qr sh twn idiot twn, na aplopoihjoÔn oi parastˆseic:
i) xy 0 + yz + z 0 w + x0 y 0 z .
ii) x(x0 + y)(z 0 + w)z .
2) Na lujoÔn oi exis¸seic thc 'Algebrac Boole:
i) x0 + xy 0 = 1.
ii) xy 0 + x0 y + yz 0 = 0.
iii) xyz 0 + x0 yz + xyz = 1.
3) Na lujoÔn ta sust mata thc 'Algebrac Boole:
½ 0
x + xy 0 = 1
i)
x + xy = 0.
½
y + x0 y + z = 1
ii)
y 0 + xz + y 0 z = 0.
½ 0
x + xy 0 + y 0 = 0
iii)
y + x 0 y + x = 1.
4) Na brejoÔn ta dÐpola pou antistoiqoÔn stic sunart seic:
i) f (x, y, z) = (x + y)z + x0 y 0 z 0 .
ii) f (x, y, z) = xyz + x0 yz + xy 0 z 0 .
12
x
y
5)
z’
y’
z
i) Na brejeÐ h sunˆrthsh pou antistoiqeÐ sto dÐpolo
x’
z’
z
x
w
y
ii) Na aplopoihjeÐ to parapˆnw dÐpolo.
6) Na aplopoihjei to parakˆtw dÐpolo.
x
x’
y
x
y
13

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