Finding the principal disjunctive normal form (PDNF) of a Boolean expression
$begingroup$
Find the principal disjunctive normal form (PDNF) of a Boolean expression
$$((pwedge q) rightarrow r)vee((pwedge q)rightarrow neg r).$$
I tried by expanding it but I am stuck with the expression
$(neg p vee neg q vee r) vee (neg p vee neg q vee neg r)$. I don't know how to convert them into min terms. Please help me.
boolean-algebra
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
$endgroup$
add a comment |
$begingroup$
Find the principal disjunctive normal form (PDNF) of a Boolean expression
$$((pwedge q) rightarrow r)vee((pwedge q)rightarrow neg r).$$
I tried by expanding it but I am stuck with the expression
$(neg p vee neg q vee r) vee (neg p vee neg q vee neg r)$. I don't know how to convert them into min terms. Please help me.
boolean-algebra
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
$endgroup$
2
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05
add a comment |
$begingroup$
Find the principal disjunctive normal form (PDNF) of a Boolean expression
$$((pwedge q) rightarrow r)vee((pwedge q)rightarrow neg r).$$
I tried by expanding it but I am stuck with the expression
$(neg p vee neg q vee r) vee (neg p vee neg q vee neg r)$. I don't know how to convert them into min terms. Please help me.
boolean-algebra
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
$endgroup$
Find the principal disjunctive normal form (PDNF) of a Boolean expression
$$((pwedge q) rightarrow r)vee((pwedge q)rightarrow neg r).$$
I tried by expanding it but I am stuck with the expression
$(neg p vee neg q vee r) vee (neg p vee neg q vee neg r)$. I don't know how to convert them into min terms. Please help me.
boolean-algebra
boolean-algebra
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
edited Jan 4 '16 at 12:09
Random Jack
2,1161519
2,1161519
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
asked Jan 3 '16 at 18:46
BalajiBalaji
70117
70117
2
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05
add a comment |
2
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05
2
2
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Expression mentioned is a Tautology.
$$((p.q) rightarrow r)+((p.q)rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$
answered Jan 9 '17 at 5:24
PriyPriy
111
$endgroup$
add a comment |
$begingroup$
The required normal form is ( if I am not wrong ):-
(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)
Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.
To convert any variable P to a minterm ,just add : ^(Qv~Q)^(Rv~R) to P
That is :P^(Qv~Q)^(Rv~R)
And then use the distributive laws.
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
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add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Expression mentioned is a Tautology.
$$((p.q) rightarrow r)+((p.q)rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$
answered Jan 9 '17 at 5:24
PriyPriy
111
$endgroup$
add a comment |
$begingroup$
Expression mentioned is a Tautology.
$$((p.q) rightarrow r)+((p.q)rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$
answered Jan 9 '17 at 5:24
PriyPriy
111
$endgroup$
add a comment |
$begingroup$
Expression mentioned is a Tautology.
$$((p.q) rightarrow r)+((p.q)rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$
answered Jan 9 '17 at 5:24
PriyPriy
111
$endgroup$
Expression mentioned is a Tautology.
$$((p.q) rightarrow r)+((p.q)rightarrow r')=((p'+q'+r)+(p'+q'+r'))=p'+q'+1 = 1$$
So PDNF corresponding to it is
$$(p'.q'.r')+(p'.q'.r)+(p'.q.r')+(p'.q.r)+(p.q'.r')+(p.q'.r)+(p.q.r')+(p.q.r)$$
answered Jan 9 '17 at 5:24
PriyPriy
111
answered Jan 9 '17 at 5:24
PriyPriy
111
answered Jan 9 '17 at 5:24
PriyPriy
111
answered Jan 9 '17 at 5:24
PriyPriy
111
111
add a comment |
add a comment |
$begingroup$
The required normal form is ( if I am not wrong ):-
(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)
Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.
To convert any variable P to a minterm ,just add : ^(Qv~Q)^(Rv~R) to P
That is :P^(Qv~Q)^(Rv~R)
And then use the distributive laws.
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
$endgroup$
add a comment |
$begingroup$
The required normal form is ( if I am not wrong ):-
(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)
Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.
To convert any variable P to a minterm ,just add : ^(Qv~Q)^(Rv~R) to P
That is :P^(Qv~Q)^(Rv~R)
And then use the distributive laws.
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
$endgroup$
add a comment |
$begingroup$
The required normal form is ( if I am not wrong ):-
(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)
Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.
To convert any variable P to a minterm ,just add : ^(Qv~Q)^(Rv~R) to P
That is :P^(Qv~Q)^(Rv~R)
And then use the distributive laws.
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
$endgroup$
The required normal form is ( if I am not wrong ):-
(p^q^r)v(~p^q^r)v(~p^~q^r)v(p^q^~r)
Once you get the form : (¬p∨¬q∨r)∨(¬p∨¬q∨¬r)
Treat this as (¬p)∨(¬q)∨(r)∨(¬p)∨(¬q)∨(¬r)
And convert the individual literals into minterms.
To convert any variable P to a minterm ,just add : ^(Qv~Q)^(Rv~R) to P
That is :P^(Qv~Q)^(Rv~R)
And then use the distributive laws.
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
edited Dec 30 '18 at 12:14
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
answered Dec 30 '18 at 12:09
Amrutesh MishraAmrutesh Mishra
11
11
add a comment |
add a comment |
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2
$begingroup$
Your last expression can be written as $text{Something}lor (rlor neg r)$.
$endgroup$
– Git Gud
Jan 3 '16 at 19:20
$begingroup$
First you must see that either the equation is sum of products or not.As it has only disjunctions, so it is not possible to find DNF for this equation.
$endgroup$
– Muhammad Asif
Dec 20 '16 at 21:05