Question

A+not(b and c)+not(not(a)+b+not(c)) ПОМОГИТЕ!

Answer 1

Базовые логические выражения: (and = +)
a) not (True) = False
b) not (False) = True
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c) True + True = True
d) True + False = False
e) False + True= False
f) False + False = False
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g) True or True = True
h) True or False = True
i) False or True = True
j) False or False = False
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В условии не сказано, кем являются A, B, C
по этому озвучим все 8 вариантов:

1) A = True; B = True; C = True

2) A = True; B = False; C = True

3) A = False; B = True; C = True

4) A = False; B = False; C = True

5) A = True; B = True; C = False

6) A = True; B = False; C = False

7) A = False; B = True; C = False

8) A = False; B = False; C = False

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и рассмотрим все 8 вариантов: (T = True; F = False)

1)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=T+not(T\ and\ T)+not(not(T)+T+not(T))=$
$=T+not(T)+not(F+T+F)=T+F+not(F)=T+F+T=F$

2)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=T+not(F\ and\ T)+not(not(T)+F+not(T))=$
$=T+not(F)+not(F+F+F)=T+T+not(F)=T+T+T=T$

3)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=F+not(T\ and\ T)+not(not(F)+T+not(T))=$
$=F+not(T)+not(T+T+F)=F+F+not(F)=F+F+T=F$

4) $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=F+not(F\ and\ T)+not(not(F)+F+not(T))=$
$=F+not(F)+not(T+F+F)=F+T+not(F)=F+T+T=F$

5)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=T+not(T\ and\ F)+not(not(T)+T+not(F))=$
$=T+not(F)+not(F+T+T)=T+T+not(F)=T+T+T=T$

6)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=T+not(F\ and\ F)+not(not(T)+F+not(F))=$
$=T+not(F)+not(F+F+T)=T+T+not(F)=T+T+T=T$

7)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=F+not(T\ and\ F)+not(not(F)+T+not(F))=$
$=F+not(F)+not(T+T+T)=F+T+not(T)=F+T+F=F$

8)  $A+not(B\ and\ C)+not(not(A)+B+not(C))=$
$=F+not(F\ and\ F)+not(not(F)+F+not(F))=$
$=F+not(F)+not(T+F+T)=F+T+not(F)=F+T+T=F$

Answer 2

To simplify the given expression A+not(b and c)+not(not(a)+b+not(c)), we can start by using De Morgan's laws and double negation to simplify the negations:

not(b and c) = not b or not c not(not(a)+b+not(c)) = (a and not b and c)

Substituting these into the original expression, we get:

A + not b or not c + (a and not b and c)

Next, we can use distributivity of "and" over "or" to simplify the last term:

A + not b or not c + a and not b

Using distributivity of "or" over "and", we can simplify the second and third terms:

not b or not c = not (b and c) a and not b = not (not a or b)

Substituting these, we get:

A + not (b and c) + not (not a or b)

Using De Morgan's laws and double negation again, we can simplify further:

not (b and c) = not b or not c not (not a or b) = a and not b

Substituting these, we get:

A + not b or not c + a and not b

Using distributivity of "and" over "or" again, we can simplify further:

not b or not c + a and not b = (not b or not c + a) and (not b or not a)

Substituting this back into the previous expression, we get:

A + (not b or not c + a) and (not b or not a)

This is the simplified form of the given expression.