logical propositions $left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$ [on hold]Logic Question : $C rightarrow(Bwedge A) = F , Alongleftrightarrow(Bwedge C) = T$ Find $Brightarrow (neg C) $proving logical equivalence $(P leftrightarrow Q) equiv (P wedge Q) vee (neg P wedge neg Q)$Showthat $mathrm p leftrightarrow mathrm q$ and $(mathrm pwedge mathrm q) vee (neg mathrm p wedge neg mathrm q)$ are logically equivalentPropositional Logic Help: $(neg p wedge (p vee q)) rightarrow q $ is a tautologyShow equivalence of statement $left(Prightarrow Qright) wedge left(Qrightarrow Rright)$ to …Logical Reasoning $(neg p vee neg q) wedge (r vee q) wedge (r implies s) implies neg(p wedge neg s)$How do you determine an interpretation and a model for $left((x wedge y) rightarrow (x vee y)right)$?Is the set $tau := leftvee, wedge, 0right$ adequate? Prove your answerlogical propositions $left ( left ( pRightarrow q right )Leftrightarrow p right )iff p wedge q$propositional logic $left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$
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logical propositions $left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$ [on hold]
Logic Question : $C rightarrow(Bwedge A) = F , Alongleftrightarrow(Bwedge C) = T$ Find $Brightarrow (neg C) $proving logical equivalence $(P leftrightarrow Q) equiv (P wedge Q) vee (neg P wedge neg Q)$Showthat $mathrm p leftrightarrow mathrm q$ and $(mathrm pwedge mathrm q) vee (neg mathrm p wedge neg mathrm q)$ are logically equivalentPropositional Logic Help: $(neg p wedge (p vee q)) rightarrow q $ is a tautologyShow equivalence of statement $left(Prightarrow Qright) wedge left(Qrightarrow Rright)$ to …Logical Reasoning $(neg p vee neg q) wedge (r vee q) wedge (r implies s) implies neg(p wedge neg s)$How do you determine an interpretation and a model for $left((x wedge y) rightarrow (x vee y)right)$?Is the set $tau := leftvee, wedge, 0right$ adequate? Prove your answerlogical propositions $left ( left ( pRightarrow q right )Leftrightarrow p right )iff p wedge q$propositional logic $left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$
$begingroup$
I need help in this exercise because I got stuck and could not get to the solution, someone could give me a clue how to continue or finish it, please(without using truth table)
$$left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$$
logic boolean-algebra
edited 2 days ago
Graham Kemp
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel
add a comment |
$begingroup$
I need help in this exercise because I got stuck and could not get to the solution, someone could give me a clue how to continue or finish it, please(without using truth table)
$$left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$$
logic boolean-algebra
edited 2 days ago
Graham Kemp
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03
add a comment |
$begingroup$
I need help in this exercise because I got stuck and could not get to the solution, someone could give me a clue how to continue or finish it, please(without using truth table)
$$left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$$
logic boolean-algebra
edited 2 days ago
Graham Kemp
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I need help in this exercise because I got stuck and could not get to the solution, someone could give me a clue how to continue or finish it, please(without using truth table)
$$left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$$
logic boolean-algebra
logic boolean-algebra
edited 2 days ago
Graham Kemp
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Graham Kemp
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Graham Kemp
86.7k43579
edited 2 days ago
Graham Kemp
86.7k43579
edited 2 days ago
Graham Kemp
86.7k43579
86.7k43579
asked Mar 10 at 21:03
lucaslucas
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 21:03
lucaslucas
63
asked Mar 10 at 21:03
lucaslucas
63
63
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lucas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel
put on hold as off-topic by Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Jyrki Lahtonen, Cesareo, Gibbs, Parcly Taxel
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03
add a comment |
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$-p lor -q$ and $q land r$ means $-p$
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$-p lor -q$ and $q land r$ means $-p$
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
add a comment |
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$-p lor -q$ and $q land r$ means $-p$
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
add a comment |
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$-p lor -q$ and $q land r$ means $-p$
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
$endgroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$-p lor -q$ and $q land r$ means $-p$
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
answered Mar 10 at 21:24
J. W. TannerJ. W. Tanner
3,2301320
3,2301320
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
add a comment |
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
i dont understand why −p∨−q and q∧r means −p ? :C
$endgroup$
– lucas
Mar 10 at 21:31
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
$begingroup$
Because $-q land q$ is a contradiction so we have $-p land q$
$endgroup$
– J. W. Tanner
Mar 10 at 22:09
add a comment |
$begingroup$
$p rightarrow -q$ means $-p lor -q$ and $(-r lor q) land r$ means $q land r$
$endgroup$
– J. W. Tanner
Mar 10 at 21:17
$begingroup$
yes, i have that in my development :D
$endgroup$
– lucas
Mar 10 at 21:24
$begingroup$
$p rightarrow -q$ is equivalent to $q rightarrow -p$ and $(-r lor q)$ is equivalent to $r rightarrow q$ If you piece these together $r$ and use Modus Ponens twice, you should have what you want.
$endgroup$
– Bernard Massé
Mar 10 at 21:26
$begingroup$
Bernard Massé if i cuold put you a vote up i would do it, but i cant so thanks you.
$endgroup$
– lucas
Mar 10 at 22:03