How to simplify this formula in DNF?Proving each conditional statement is a tautologyUse tableau to convert formula to DNF/CNF formDNF simplificationfinding a formula for a given truth tableConverting DNF to CNF and vice versaProving Truth Tables and FormulasFor a given truth function, how can we express it in its simplest form?Simplifying a logical equivalenceHow do you determine an interpretation and a model for $left((x wedge y) rightarrow (x vee y)right)$?Prove that every to $varphi_n$ equivalent formula in DNF has at least $2^n$ conjunctive clauses
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How to simplify this formula in DNF?
Proving each conditional statement is a tautologyUse tableau to convert formula to DNF/CNF formDNF simplificationfinding a formula for a given truth tableConverting DNF to CNF and vice versaProving Truth Tables and FormulasFor a given truth function, how can we express it in its simplest form?Simplifying a logical equivalenceHow do you determine an interpretation and a model for $left((x wedge y) rightarrow (x vee y)right)$?Prove that every to $varphi_n$ equivalent formula in DNF has at least $2^n$ conjunctive clauses
$begingroup$
Given a truth table, a formula in DNF was created, as shown below:
(-p ^ -q ^ r) v (p ^ -q ^ r) v (p ^ q ^ -r) v (p ^ q ^ r)
How would the simplification be represented?
There are two r's, and two q's. Can the distributive property be used? A step-by-step explanation would be greatly appreciated. Thank you.
propositional-calculus disjunctive-normal-form
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
$endgroup$
add a comment |
$begingroup$
Given a truth table, a formula in DNF was created, as shown below:
(-p ^ -q ^ r) v (p ^ -q ^ r) v (p ^ q ^ -r) v (p ^ q ^ r)
How would the simplification be represented?
There are two r's, and two q's. Can the distributive property be used? A step-by-step explanation would be greatly appreciated. Thank you.
propositional-calculus disjunctive-normal-form
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
$endgroup$
$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19
add a comment |
$begingroup$
Given a truth table, a formula in DNF was created, as shown below:
(-p ^ -q ^ r) v (p ^ -q ^ r) v (p ^ q ^ -r) v (p ^ q ^ r)
How would the simplification be represented?
There are two r's, and two q's. Can the distributive property be used? A step-by-step explanation would be greatly appreciated. Thank you.
propositional-calculus disjunctive-normal-form
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
$endgroup$
Given a truth table, a formula in DNF was created, as shown below:
(-p ^ -q ^ r) v (p ^ -q ^ r) v (p ^ q ^ -r) v (p ^ q ^ r)
How would the simplification be represented?
There are two r's, and two q's. Can the distributive property be used? A step-by-step explanation would be greatly appreciated. Thank you.
propositional-calculus disjunctive-normal-form
propositional-calculus disjunctive-normal-form
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
edited Mar 22 at 3:11
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Mar 20 at 18:58
SaraSara
245
asked Mar 20 at 18:58
SaraSara
245
asked Mar 20 at 18:58
SaraSara
245
245
$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19
add a comment |
$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19
$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19
$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19
add a comment |
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$begingroup$
Yes, use Distributivity : the part $(lnot p land (lnot q land r)) lor (p land (lnot q land r))$ is equiv to $((p lor lnot p) land (lnot q land r))$. In turn this is $(text T land (lnot q land r)) equiv (lnot q land r)$.
$endgroup$
– Mauro ALLEGRANZA
Mar 22 at 7:19