Algebra booleana

CURSO: SISTEMA DE INFORMAÇÃO DISCIPLINA: ORGANIZAÇÃO DE COMPUTADORES PROFº: PÉRICLES 1º SEMESTRE

ÁLGEBRA BOOLEANA

ANTONIO ELLISON CARLOS TAVARES

201301501727 000000000000

FORTALEZA 2013

1ª QUESTÃO a) A 1 1 1 0 0 0 1 0 B 1 1 0 1 0 1 0 0 C 1 0 1 1 1 0 0 0 A.B.C+¬A.¬B.¬C 1 0 0 0 0 0 0 1

b) A 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 C 1 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 B 1 1 0 1 1 0 1 0 0 0 0 1 0 1 1 0 D 1 0 1 1 1 1 0 0 0 0 0 1 1 0 0 1 A∧(C∨∼B∨∼D) 1 1 1 0 0 0 0 0 1 0 1 0 1 0 1 0

c) A 1 1 1 0 0 0 1 0 B 1 1 0 1 0 1 0 0 C 1 0 1 1 1 0 0 0 A∧B∧C∨A∧∼B∧∼C∨∼A∧∼B∧∼C 1 0 0 0 0 0 1 1

d) A 1 1 1 0 0 0 1 0 B 1 1 0 1 0 1 0 0 C 1 0 1 1 1 0 0 0 (A∨B)∧(∼A∨∼C)∧(∼A B) 0 1 0 0 0 0 0 0

e) A 0 0 1 1 B 1 0 1 0 A∧B∨A∧∼B 0 0 1 1

f) A 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 B 1 1 1 0 1 0 0 1 0 0 1 0 0 1 0 1 C 1 1 0 1 1 0 1 0 0 0 0 1 0 1 1 0 D 1 0 1 1 1 1 0 0 0 0 0 1 1 0 0 1 A∨∼(∼B∨A∧C) 1 0 1 1 1 0 1 0 0 1 0 0 1 0 0 1 ∼D

2ª QUESTÃO A) X= A.(B C) 1011. (1110 0011) = 1011.1101 = 1001 B) X= ¬ (A+B) . ( C (A+¬D)) ¬ (1011 + 1110) ¬ (1111) 0000 . (0011 (1011+¬1010)) 0000 C) X= B .¬ C . A + ¬(¬C D)

1110.1100.1011+¬(1100 1010) 1100.1011+1001 1000+1001 1001

D) X=((¬A+¬(B ¬D)).(C+A)+B).¬(A+B) ((0100+¬1110 0101).(0011+1011)+1110).¬(1011+1110) 0100+¬1011.1011+1110.¬1111 0100+0100.1111.0000 0100.0000 0000 E) X=A B+¬C.B.¬A 10011 1110+1100.1110+0100 0101+1100+0100 0101+1100 1100

3ª QUESTÃO A) A 1 1 1 0 0 0 1 0 B 1 1 0 1 0 1 0 0 C 1 0 1 1 1 0 0 0 (X∨Y)∨Q 1 1 1 1 1 1 1 0

B) X 1 1 0 0 Y 1 0 1 0 (X∧Y)∨Y 1 0 1 0