Boolean expression simplified

-1

$begingroup$

c. Interpret the following complex Boolean expression and produce a truth table and a logic circuit?

---

---

(x+y•x) + (x•y)

.=AND
+=or

boolean-algebra

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edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

$endgroup$

add a comment | 

-1

$begingroup$

c. Interpret the following complex Boolean expression and produce a truth table and a logic circuit?

---

---

(x+y•x) + (x•y)

.=AND
+=or

boolean-algebra

share|cite|improve this question

edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

$endgroup$

add a comment | 

-1

-1

-1

$begingroup$

c. Interpret the following complex Boolean expression and produce a truth table and a logic circuit?

---

---

(x+y•x) + (x•y)

.=AND
+=or

boolean-algebra

share|cite|improve this question

edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

$endgroup$

c. Interpret the following complex Boolean expression and produce a truth table and a logic circuit?

---

---

(x+y•x) + (x•y)

.=AND
+=or

boolean-algebra

boolean-algebra

share|cite|improve this question

edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

share|cite|improve this question

edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

share|cite|improve this question

share|cite|improve this question

edited Jan 23 at 13:14

Callum Smith

edited Jan 23 at 13:14

Callum Smith

edited Jan 23 at 13:14

Callum Smith

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

asked Jan 23 at 10:14

Callum SmithCallum Smith

11

11

add a comment | 

add a comment | 

1 Answer
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If the dot is "and" then $(x+yx) +xy stackrel{xy=yx}{=} x(1+y)+xy stackrel{1+y=1}{=} x+xy = x(1+y) = x$

share|cite|improve this answer

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

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$begingroup$

If the dot is "and" then $(x+yx) +xy stackrel{xy=yx}{=} x(1+y)+xy stackrel{1+y=1}{=} x+xy = x(1+y) = x$

share|cite|improve this answer

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

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0

$begingroup$

If the dot is "and" then $(x+yx) +xy stackrel{xy=yx}{=} x(1+y)+xy stackrel{1+y=1}{=} x+xy = x(1+y) = x$

share|cite|improve this answer

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

$endgroup$

add a comment | 

0

0

0

$begingroup$

If the dot is "and" then $(x+yx) +xy stackrel{xy=yx}{=} x(1+y)+xy stackrel{1+y=1}{=} x+xy = x(1+y) = x$

share|cite|improve this answer

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

$endgroup$

If the dot is "and" then $(x+yx) +xy stackrel{xy=yx}{=} x(1+y)+xy stackrel{1+y=1}{=} x+xy = x(1+y) = x$

share|cite|improve this answer

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

share|cite|improve this answer

share|cite|improve this answer

edited Jan 23 at 11:36

edited Jan 23 at 11:36

edited Jan 23 at 11:36

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

answered Jan 23 at 10:19

trancelocationtrancelocation

12.6k1826

12.6k1826

add a comment | 

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