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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Algèbre de BooleLogique de base/Exercices/Algèbre de Boole », n'a pu être restituée correctement ci-dessus.
Réduire les équations en utilisant les propriétés de l'algèbre de Boole:
A
=
a
+
a
⋅
b
{\displaystyle A=a+a\cdot b}
Solution
Je factorise par rapport à a dans tous les termes:
A
=
a
(
1
+
b
)
=
a
⋅
1
=
a
{\displaystyle {\begin{aligned}A&=a(1+b)\\&=a\cdot 1\\&=a\\\end{aligned}}}
A
=
a
+
a
⋅
b
=
a
{\displaystyle A=a+a\cdot b=a}
B
=
a
⋅
(
a
+
b
)
{\displaystyle B=a\cdot (a+b)}
Solution
Je développe
a
{\displaystyle a}
B
=
a
⋅
a
+
a
⋅
b
=
a
+
a
⋅
b
{\displaystyle {\begin{aligned}B&=a\cdot a+a\cdot b\\&=a+a\cdot b\end{aligned}}}
Je factorise par rapport à
a
{\displaystyle a}
B
=
a
⋅
(
1
+
b
)
=
a
⋅
1
=
a
{\displaystyle {\begin{aligned}B&=a\cdot (1+b)\\&=a\cdot 1\\&=a\end{aligned}}}
B
=
a
⋅
(
a
+
b
)
=
a
{\displaystyle B=a\cdot (a+b)=a}
C
=
a
+
(
a
¯
⋅
b
)
{\displaystyle C=a+({\bar {a}}\cdot b)}
Solution
C
=
(
a
+
a
¯
)
⋅
(
a
+
b
)
=
1
⋅
(
a
+
b
)
=
a
+
b
{\displaystyle {\begin{aligned}C&=(a+{\bar {a}})\cdot (a+b)\\&=1\cdot (a+b)\\&=a+b\end{aligned}}}
C
=
a
+
(
a
¯
⋅
b
)
=
a
+
b
{\displaystyle C=a+({\bar {a}}\cdot b)=a+b}
D
=
(
a
+
b
)
⋅
(
a
+
b
¯
)
{\displaystyle D=(a+b)\cdot (a+{\bar {b}})}
Solution
Je développe la première parenthèse avec la deuxième :
D
=
a
⋅
a
+
a
⋅
b
¯
+
b
⋅
a
+
b
⋅
b
¯
=
a
+
a
⋅
b
¯
+
a
⋅
b
+
0
=
a
+
a
⋅
b
¯
+
a
⋅
b
{\displaystyle {\begin{aligned}D&=a\cdot a+a\cdot {\bar {b}}+b\cdot a+b\cdot {\bar {b}}\\&=a+a\cdot {\bar {b}}+a\cdot b+0\\&=a+a\cdot {\bar {b}}+a\cdot b\end{aligned}}}
Je factorise par rapport à a dans tous les termes:
D
=
a
(
1
+
b
¯
+
b
)
=
a
⋅
1
=
a
{\displaystyle {\begin{aligned}D&=a(1+{\bar {b}}+b)\\&=a\cdot 1\\&=a\end{aligned}}}
D
=
(
a
+
b
)
⋅
(
a
+
b
¯
)
=
a
{\displaystyle D=(a+b)\cdot (a+{\bar {b}})=a}
G
=
(
a
¯
+
b
¯
)
+
(
a
¯
+
b
)
+
(
a
+
b
¯
)
{\displaystyle G=({\bar {a}}+{\bar {b}})+({\bar {a}}+b)+(a+{\bar {b}})}
Solution
On retire les parenthèses
G
=
a
¯
+
b
¯
+
a
¯
+
b
+
a
+
b
¯
{\displaystyle {\begin{aligned}G&={\bar {a}}+{\bar {b}}+{\bar {a}}+b+a+{\bar {b}}\\\end{aligned}}}
a et son complément valent 1 et 1+ n'importe quelle valeur vaut 1
G
=
1
{\displaystyle {\begin{aligned}G&=1\end{aligned}}}
H
=
(
a
¯
+
b
)
+
(
a
+
b
¯
)
+
(
a
+
b
)
{\displaystyle H=({\bar {a}}+b)+(a+{\bar {b}})+(a+b)}
E
=
(
a
+
b
)
⋅
(
a
+
c
)
{\displaystyle E=(a+b)\cdot (a+c)}
Solution
Je développe la première parenthèse avec la deuxième :
E
=
a
⋅
a
+
a
⋅
c
+
b
⋅
a
+
b
⋅
c
=
a
+
a
⋅
c
+
a
⋅
b
+
b
⋅
c
{\displaystyle {\begin{aligned}E&=a\cdot a+a\cdot c+b\cdot a+b\cdot c\\&=a+a\cdot c+a\cdot b+b\cdot c\end{aligned}}}
Je factorise a dans tous les termes, sauf dans le dernier :
E
=
a
(
1
+
c
+
b
)
+
b
⋅
c
=
a
.1
+
b
⋅
c
=
a
+
b
⋅
c
{\displaystyle {\begin{aligned}E&=a(1+c+b)+b\cdot c\\&=a.1+b\cdot c\\&=a+b\cdot c\end{aligned}}}
E
=
(
a
+
b
)
⋅
(
a
+
c
)
=
a
+
b
⋅
c
{\displaystyle E=(a+b)\cdot (a+c)=a+b\cdot c}
F
=
(
a
+
b
)
⋅
(
a
¯
+
c
)
{\displaystyle F=(a+b)\cdot ({\bar {a}}+c)}
Solution
Je développe la première parenthèse avec la deuxième :
F
=
a
⋅
a
¯
+
a
⋅
c
+
b
⋅
a
¯
+
b
⋅
c
=
0
+
a
⋅
c
+
a
¯
⋅
b
+
b
⋅
c
{\displaystyle {\begin{aligned}F&=a\cdot {\bar {a}}+a\cdot c+b\cdot {\bar {a}}+b\cdot c\\&=0+a\cdot c+{\bar {a}}\cdot b+b\cdot c\\\end{aligned}}}
Théorème du consensus
F
=
a
¯
⋅
b
+
a
⋅
c
{\displaystyle {\begin{aligned}F&={\bar {a}}\cdot b+a\cdot c\end{aligned}}}
P
=
(
a
+
b
¯
)
⋅
(
b
+
c
¯
)
⋅
(
c
+
a
¯
)
{\displaystyle P=(a+{\bar {b}})\cdot (b+{\bar {c}})\cdot (c+{\bar {a}})}
Solution
On développe la première parenthèse avec la deuxième :
P
=
(
a
⋅
b
+
a
⋅
c
¯
+
b
¯
⋅
b
+
b
¯
⋅
c
¯
)
⋅
(
c
+
a
¯
)
=
(
a
⋅
b
+
a
⋅
c
¯
+
0
+
b
¯
⋅
c
¯
)
⋅
(
c
+
a
¯
)
=
(
a
⋅
b
+
a
⋅
c
¯
+
b
¯
⋅
c
¯
)
⋅
(
c
+
a
¯
)
{\displaystyle {\begin{aligned}P&=(a\cdot b+a\cdot {\bar {c}}+{\bar {b}}\cdot b+{\bar {b}}\cdot {\bar {c}})\cdot (c+{\bar {a}})\\&=(a\cdot b+a\cdot {\bar {c}}+0+{\bar {b}}\cdot {\bar {c}})\cdot (c+{\bar {a}})\\&=(a\cdot b+a\cdot {\bar {c}}+{\bar {b}}\cdot {\bar {c}})\cdot (c+{\bar {a}})\end{aligned}}}
On développe la première parenthèse avec la deuxième :
P
=
a
⋅
b
⋅
c
+
a
⋅
c
¯
⋅
c
+
b
¯
⋅
c
¯
⋅
c
+
a
⋅
b
⋅
a
¯
+
a
⋅
c
¯
⋅
a
¯
+
b
¯
⋅
c
¯
⋅
a
¯
=
a
⋅
b
⋅
c
+
a
.0
+
b
¯
.0
+
a
⋅
a
¯
⋅
b
+
a
⋅
a
¯
⋅
c
¯
+
b
¯
⋅
c
¯
⋅
a
¯
=
a
⋅
b
⋅
c
+
0
+
0
+
0
⋅
b
+
0
⋅
c
¯
+
a
¯
⋅
b
¯
⋅
c
¯
=
a
⋅
b
⋅
c
+
0
+
0
+
0
+
0
+
a
¯
⋅
b
¯
⋅
c
¯
=
a
⋅
b
⋅
c
+
a
¯
⋅
b
¯
⋅
c
¯
{\displaystyle {\begin{aligned}P&=a\cdot b\cdot c+a\cdot {\bar {c}}\cdot c+{\bar {b}}\cdot {\bar {c}}\cdot c+a\cdot b\cdot {\bar {a}}+a\cdot {\bar {c}}\cdot {\bar {a}}+{\bar {b}}\cdot {\bar {c}}\cdot {\bar {a}}\\&=a\cdot b\cdot c+a.0+{\bar {b}}.0+a\cdot {\bar {a}}\cdot b+a\cdot {\bar {a}}\cdot {\bar {c}}+{\bar {b}}\cdot {\bar {c}}\cdot {\bar {a}}\\&=a\cdot b\cdot c+0+0+0\cdot b+0\cdot {\bar {c}}+{\bar {a}}\cdot {\bar {b}}\cdot {\bar {c}}\\&=a\cdot b\cdot c+0+0+0+0+{\bar {a}}\cdot {\bar {b}}\cdot {\bar {c}}\\&=a\cdot b\cdot c+{\bar {a}}\cdot {\bar {b}}\cdot {\bar {c}}\end{aligned}}}
P
=
a
⋅
b
⋅
c
+
a
¯
.
b
¯
.
c
¯
{\displaystyle P=a\cdot b\cdot c+{\bar {a}}.{\bar {b}}.{\bar {c}}}
I
=
(
b
¯
+
a
¯
)
⋅
(
a
⋅
c
+
b
¯
)
{\displaystyle I=({\bar {b}}+{\bar {a}})\cdot (a\cdot c+{\bar {b}})}
Solution
I
=
a
⋅
b
¯
⋅
c
+
a
⋅
a
¯
⋅
c
+
b
¯
⋅
b
¯
+
a
¯
⋅
b
¯
=
a
⋅
b
¯
⋅
c
+
b
¯
+
a
¯
⋅
b
¯
=
b
¯
⋅
(
a
⋅
c
+
1
+
a
)
=
b
¯
⋅
1
I
=
b
¯
{\displaystyle {\begin{aligned}I&=a\cdot {\bar {b}}\cdot c+a\cdot {\bar {a}}\cdot c+{\bar {b}}\cdot {\bar {b}}+{\bar {a}}\cdot {\bar {b}}\\&=a\cdot {\bar {b}}\cdot c+{\bar {b}}+{\bar {a}}\cdot {\bar {b}}\\&={\bar {b}}\cdot (a\cdot c+1+a)\\&={\bar {b}}\cdot 1\\I&={\bar {b}}\end{aligned}}}
Q
=
(
a
+
b
+
c
)
⋅
(
a
+
b
¯
+
c
)
⋅
(
a
+
b
¯
+
c
¯
)
{\displaystyle Q=(a+b+c)\cdot (a+{\bar {b}}+c)\cdot (a+{\bar {b}}+{\bar {c}})}
Solution
Je développe les deux premières parenthèses :
Q
=
(
a
⋅
a
+
a
⋅
b
¯
+
a
⋅
c
+
b
⋅
a
+
b
⋅
b
¯
+
b
⋅
c
+
c
⋅
a
+
c
⋅
b
¯
+
c
⋅
c
)
⋅
(
a
+
b
¯
+
c
¯
)
=
(
a
+
a
⋅
b
¯
+
a
⋅
c
+
a
⋅
b
+
0
+
b
⋅
c
+
a
⋅
c
+
b
¯
⋅
c
+
c
)
⋅
(
a
+
b
¯
+
c
¯
)
=
(
a
+
a
⋅
b
¯
+
a
⋅
c
+
a
⋅
b
+
b
⋅
c
+
a
⋅
c
+
b
¯
⋅
c
+
c
)
⋅
(
a
+
b
¯
+
c
¯
)
=
(
a
+
a
⋅
b
¯
+
a
⋅
c
+
a
⋅
b
+
b
⋅
c
+
b
¯
⋅
c
+
c
)
⋅
(
a
+
b
¯
+
c
¯
)
{\displaystyle {\begin{aligned}Q&=(a\cdot a+a\cdot {\bar {b}}+a\cdot c+b\cdot a+b\cdot {\bar {b}}+b\cdot c+c\cdot a+c\cdot {\bar {b}}+c\cdot c)\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a+a\cdot {\bar {b}}+a\cdot c+a\cdot b+0+b\cdot c+a\cdot c+{\bar {b}}\cdot c+c)\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a+a\cdot {\bar {b}}+a\cdot c+a\cdot b+b\cdot c+a\cdot c+{\bar {b}}\cdot c+c)\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a+a\cdot {\bar {b}}+a\cdot c+a\cdot b+b\cdot c+{\bar {b}}\cdot c+c)\cdot (a+{\bar {b}}+{\bar {c}})\end{aligned}}}
Dans la première parenthèse, je factorise
a
{\displaystyle a}
c
{\displaystyle c}
Q
=
[
a
⋅
(
1
+
b
¯
+
c
+
b
)
+
c
⋅
(
b
+
b
¯
+
1
)
]
⋅
(
a
+
b
¯
+
c
¯
)
=
(
a
.1
+
c
.1
)
⋅
(
a
+
b
¯
+
c
¯
)
=
(
a
+
c
)
⋅
(
a
+
b
¯
+
c
¯
)
=
a
⋅
a
+
a
⋅
b
¯
+
a
⋅
c
¯
+
c
⋅
a
+
c
⋅
b
¯
+
c
⋅
c
¯
=
a
+
a
⋅
b
¯
+
a
⋅
c
¯
+
a
⋅
c
+
b
¯
⋅
c
+
0
=
a
+
a
⋅
b
¯
+
a
⋅
c
¯
+
a
⋅
c
+
b
¯
⋅
c
{\displaystyle {\begin{aligned}Q&=[a\cdot (1+{\bar {b}}+c+b)+c\cdot (b+{\bar {b}}+1)]\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a.1+c.1)\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a+c)\cdot (a+{\bar {b}}+{\bar {c}})\\&=a\cdot a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+c\cdot a+c\cdot {\bar {b}}+c\cdot {\bar {c}}\\&=a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+a\cdot c+{\bar {b}}\cdot c+0\\&=a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+a\cdot c+{\bar {b}}\cdot c\end{aligned}}}
Je factorise
a
{\displaystyle a}
Q
=
a
.
(
1
+
b
¯
+
c
¯
+
c
)
+
b
¯
⋅
c
=
a
.1
+
b
¯
⋅
c
=
a
+
b
¯
⋅
c
{\displaystyle {\begin{aligned}Q&=a.(1+{\bar {b}}+{\bar {c}}+c)+{\bar {b}}\cdot c\\&=a.1+{\bar {b}}\cdot c\\&=a+{\bar {b}}\cdot c\end{aligned}}}
Q
=
a
+
b
¯
.
c
{\displaystyle Q=a+{\bar {b}}.c}
T
=
a
⋅
b
⋅
c
+
a
⋅
b
¯
⋅
c
+
a
⋅
b
⋅
c
¯
{\displaystyle T=a\cdot b\cdot c+a\cdot {\bar {b}}\cdot c+a\cdot b\cdot {\bar {c}}}
Solution
On factorise a dans les 3 termes :
T
=
a
⋅
(
b
⋅
c
+
b
¯
⋅
c
+
b
⋅
c
¯
)
{\displaystyle T=a\cdot (b\cdot c+{\bar {b}}\cdot c+b\cdot {\bar {c}})}
Dans la parenthèse, on factorise dans le premier et le dernier terme :
T
=
a
⋅
[
b
⋅
(
c
+
c
¯
)
+
b
¯
⋅
c
]
=
a
⋅
[
b
⋅
1
+
b
¯
⋅
c
]
=
a
⋅
(
b
+
b
¯
⋅
c
)
=
a
⋅
[
(
b
+
b
¯
)
⋅
(
b
+
c
)
]
=
a
⋅
[
1
⋅
(
b
+
c
)
]
=
a
⋅
(
b
+
c
)
{\displaystyle {\begin{aligned}T&=a\cdot [b\cdot (c+{\bar {c}})+{\bar {b}}\cdot c]\\&=a\cdot [b\cdot 1+{\bar {b}}\cdot c]\\&=a\cdot (b+{\bar {b}}\cdot c)\\&=a\cdot [(b+{\bar {b}})\cdot (b+c)]\\&=a\cdot [1\cdot (b+c)]\\&=a\cdot (b+c)\end{aligned}}}
T
=
a
⋅
(
b
+
c
)
{\displaystyle T=a\cdot (b+c)}
V
=
(
a
+
b
)
⋅
(
a
+
c
)
+
(
b
+
c
)
⋅
(
b
+
a
)
+
(
c
+
a
)
⋅
(
c
+
b
)
{\displaystyle V=(a+b)\cdot (a+c)+(b+c)\cdot (b+a)+(c+a)\cdot (c+b)}
Solution
Je développe les 3 groupes de 2 parenthèses chacune :
V
=
a
⋅
a
+
a
⋅
c
+
b
⋅
a
+
b
⋅
c
+
b
⋅
b
+
b
⋅
a
+
b
⋅
c
+
c
⋅
a
+
c
⋅
c
+
c
⋅
b
+
a
⋅
c
+
a
⋅
b
=
a
+
a
⋅
c
+
a
⋅
b
+
b
⋅
c
+
b
+
a
⋅
b
+
b
⋅
c
+
a
⋅
c
+
c
+
b
⋅
c
+
a
⋅
c
+
a
⋅
b
=
a
+
a
⋅
c
+
a
⋅
b
+
b
⋅
c
+
b
+
c
{\displaystyle {\begin{aligned}V&=a\cdot a+a\cdot c+b\cdot a+b\cdot c+b\cdot b+b\cdot a+b\cdot c+c\cdot a+c\cdot c+c\cdot b+a\cdot c+a\cdot b\\&=a+a\cdot c+a\cdot b+b\cdot c+b+a\cdot b+b\cdot c+a\cdot c+c+b\cdot c+a\cdot c+a\cdot b\\&=a+a\cdot c+a\cdot b+b\cdot c+b+c\end{aligned}}}
Je factorise a dans les 3 premiers termes et b dans les 2 termes suivants :
V
=
a
(
1
+
c
+
b
)
+
b
(
c
+
1
)
+
c
=
a
.1
+
b
.1
+
c
=
a
+
b
+
c
{\displaystyle {\begin{aligned}V&=a(1+c+b)+b(c+1)+c\\&=a.1+b.1+c\\&=a+b+c\end{aligned}}}
V
=
a
+
b
+
c
{\displaystyle V=a+b+c}
K
=
a
⋅
c
⋅
(
a
¯
+
b
+
c
¯
)
{\displaystyle K=a\cdot c\cdot ({\bar {a}}+b+{\bar {c}})}
Solution
On développe l'expression :
K
=
a
⋅
a
¯
⋅
c
+
a
⋅
b
⋅
c
+
a
⋅
c
⋅
c
¯
{\displaystyle {\begin{aligned}K&=a\cdot {\bar {a}}\cdot c+a\cdot b\cdot c+a\cdot c\cdot {\bar {c}}\end{aligned}}}
Les premiers et derniers produits disparaissent, d'où :
K
=
a
⋅
b
⋅
c
{\displaystyle K=a\cdot b\cdot c}
L
=
a
b
+
a
¯
b
+
a
c
+
a
¯
c
{\displaystyle L=ab+{\bar {a}}b+ac+{\bar {a}}c}
Solution
On factorise par b et par c :
L
=
b
⋅
(
a
+
a
¯
)
+
c
⋅
(
a
+
a
¯
)
=
b
⋅
1
+
c
⋅
1
{\displaystyle {\begin{aligned}L&=b\cdot (a+{\bar {a}})+c\cdot (a+{\bar {a}})\\&=b\cdot 1+c\cdot 1\end{aligned}}}
L
=
b
+
c
{\displaystyle L=b+c}
U
=
(
a
¯
+
b
)
⋅
(
a
+
b
+
d
)
⋅
d
¯
{\displaystyle U=({\bar {a}}+b)\cdot (a+b+d)\cdot {\bar {d}}}
Solution
Je développe la dernière parenthèse avec
d
¯
{\displaystyle {\bar {d}}}
U
=
(
a
¯
+
b
)
⋅
(
a
⋅
d
¯
+
b
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{\displaystyle {\begin{aligned}U&=({\bar {a}}+b)\cdot (a\cdot {\bar {d}}+b\cdot {\bar {d}}+d\cdot {\bar {d}})\\&=({\bar {a}}+b)\cdot (a\cdot {\bar {d}}+b\cdot {\bar {d}}+0)\\&=({\bar {a}}+b)\cdot (a\cdot {\bar {d}}+b\cdot {\bar {d}})\end{aligned}}}
Je développe les parenthèses :
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{\displaystyle {\begin{aligned}U&={\bar {a}}\cdot a\cdot {\bar {d}}+{\bar {a}}\cdot b\cdot {\bar {d}}+b\cdot a\cdot {\bar {d}}+b\cdot b\cdot {\bar {d}}\\&=0\cdot {\bar {d}}+{\bar {a}}\cdot b\cdot {\bar {d}}+b\cdot a\cdot {\bar {d}}+b\cdot {\bar {d}}\\&={\bar {a}}\cdot b\cdot {\bar {d}}+a\cdot b\cdot {\bar {d}}+b\cdot {\bar {d}}\end{aligned}}}
Je factorise
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{\displaystyle {\begin{aligned}b\cdot {\bar {d}}\end{aligned}}}
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{\displaystyle {\begin{aligned}U&=(b\cdot {\bar {d}})\cdot ({\bar {a}}+a+1)\\&=(b\cdot {\bar {d}})\cdot 1\\&=b\cdot {\bar {d}}\end{aligned}}}
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{\displaystyle U=b\cdot {\bar {d}}}
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{\displaystyle Z=(a\cdot b+c+d)\cdot a\cdot b}
Solution
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{\displaystyle Z=(a\cdot b+c+d)\cdot a\cdot b}
Je développe les parenthèses :
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{\displaystyle {\begin{aligned}Z&=a\cdot a\cdot b\cdot b+a\cdot b\cdot c+a\cdot b\cdot d\\&=a\cdot b+a\cdot b\cdot c+a\cdot b\cdot d\end{aligned}}}
Je factorise
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{\displaystyle {\begin{aligned}a\cdot b\end{aligned}}}
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{\displaystyle {\begin{aligned}Z&=(a\cdot b)\cdot (1+c+d)\\&=(a\cdot b)\cdot 1\\&=a\cdot b\end{aligned}}}
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{\displaystyle Z=a\cdot b}
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{\displaystyle J=c\cdot (b+c)+(a+d)\cdot ({\bar {a}}+d)\cdot {\bar {c}}}
Solution
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{\displaystyle J=c\cdot (b+c)+(a+d)\cdot ({\bar {a}}+d)\cdot {\bar {c}}}
Je développe les parenthèses :
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{\displaystyle {\begin{aligned}J&=c\cdot b+c\cdot c+a\cdot {\bar {a}}\cdot {\bar {c}}+a\cdot d\cdot {\bar {c}}+d\cdot {\bar {a}}\cdot {\bar {c}}+d\cdot d\cdot {\bar {c}}\\&=c\cdot b+c+0+a\cdot d\cdot {\bar {c}}+d\cdot {\bar {a}}\cdot {\bar {c}}+d\cdot {\bar {c}}\end{aligned}}}
Je factorise par
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{\displaystyle c}
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{\displaystyle d{\bar {c}}}
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{\displaystyle {\begin{aligned}J&=(b+1)\cdot c+(a+{\bar {a}}+1)d\cdot {\bar {c}}\\&=c+d\cdot {\bar {c}}\\&=(1+d)\cdot c+d\cdot {\bar {c}}\\&=c+dc+d\cdot {\bar {c}}\\&=c+d(c+{\bar {c}})\\&=c+d\times 1\\&=c+d\\\end{aligned}}}
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{\displaystyle J=c+d}
![{\displaystyle {\begin{aligned}Q&=[a\cdot (1+{\bar {b}}+c+b)+c\cdot (b+{\bar {b}}+1)]\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a.1+c.1)\cdot (a+{\bar {b}}+{\bar {c}})\\&=(a+c)\cdot (a+{\bar {b}}+{\bar {c}})\\&=a\cdot a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+c\cdot a+c\cdot {\bar {b}}+c\cdot {\bar {c}}\\&=a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+a\cdot c+{\bar {b}}\cdot c+0\\&=a+a\cdot {\bar {b}}+a\cdot {\bar {c}}+a\cdot c+{\bar {b}}\cdot c\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d3b996367a7d9f14dadc9f5ae7f86e157662338)
![{\displaystyle {\begin{aligned}T&=a\cdot [b\cdot (c+{\bar {c}})+{\bar {b}}\cdot c]\\&=a\cdot [b\cdot 1+{\bar {b}}\cdot c]\\&=a\cdot (b+{\bar {b}}\cdot c)\\&=a\cdot [(b+{\bar {b}})\cdot (b+c)]\\&=a\cdot [1\cdot (b+c)]\\&=a\cdot (b+c)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f642eb9318114826a8172fa0aa2862333dc5e5d)
