propositional logic $left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$ [on...
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propositional logic $left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$ [on hold]
Logic Question : $C rightarrow(Bwedge A) = F , Alongleftrightarrow(Bwedge C) = T$ Find $Brightarrow (neg C) $Propositional Logic Help: $(neg p wedge (p vee q)) rightarrow q $ is a tautologyNeed Help with Propositional LogicShow equivalence of statement $left(Prightarrow Qright) wedge left(Qrightarrow Rright)$ to …Solving Logical equivalence & propositional logic problems without truth tablesProving that $(neg Q) Rightarrow (R Rightarrow neg (P wedge Q))$ is a tautology, by contradictionHow do you determine an interpretation and a model for $left((x wedge y) rightarrow (x vee y)right)$?Is the set $tau := left{vee, wedge, 0right}$ adequate? Prove your answerlogical propositions $left ( left ( pRightarrow q right )Leftrightarrow p right )iff p wedge q$logical propositions $left [ left ( prightarrow -q right ) wedge left ( -rvee q right )wedge rright ]rightarrow -p$
$begingroup$
I need help in this problem, I got stuck and I can not solve it, please help me with an hint(without using truth table)
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]rightarrow left [ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
logic boolean-algebra
asked Mar 10 at 22:34
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 Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra Mar 11 at 10:20
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." – Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
I need help in this problem, I got stuck and I can not solve it, please help me with an hint(without using truth table)
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]rightarrow left [ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
logic boolean-algebra
asked Mar 10 at 22:34
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 Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra Mar 11 at 10:20
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." – Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59
|
show 3 more comments
$begingroup$
I need help in this problem, I got stuck and I can not solve it, please help me with an hint(without using truth table)
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]rightarrow left [ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
logic boolean-algebra
asked Mar 10 at 22:34
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 problem, I got stuck and I can not solve it, please help me with an hint(without using truth table)
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]rightarrow left [ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
logic boolean-algebra
logic boolean-algebra
asked Mar 10 at 22:34
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 22:34
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 Mar 10 at 22:43
lucas
asked Mar 10 at 22:34
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 22:34
lucaslucas
63
asked Mar 10 at 22:34
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 Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra Mar 11 at 10:20
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." – Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra Mar 11 at 10:20
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." – Ennar, Parcly Taxel, Jyrki Lahtonen, Vinyl_cape_jawa, Riccardo.Alestra
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59
|
show 3 more comments
1
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59
1
1
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If the left-hand side of your main implication, i. e.
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$$
is true, then there are 2 possibilities:
Both $p$ and $r$ are true . Then both $q$ and $s$ must be true, too.
The right-hand side of your implication is then true, because it is reduced to "true $rightarrow$ true".
At least one of $p, r$ is false, i. e. $(p wedge r)$ is false. Then the right-hand side of your implication
$$left[ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
is true, because the implication "false $rightarrow$ anything" is always true.
If the left-hand side of your main implication is false, then there is nothing to prove, because - again - the implication "false $rightarrow$ anything" is always true.
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the left-hand side of your main implication, i. e.
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$$
is true, then there are 2 possibilities:
Both $p$ and $r$ are true . Then both $q$ and $s$ must be true, too.
The right-hand side of your implication is then true, because it is reduced to "true $rightarrow$ true".
At least one of $p, r$ is false, i. e. $(p wedge r)$ is false. Then the right-hand side of your implication
$$left[ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
is true, because the implication "false $rightarrow$ anything" is always true.
If the left-hand side of your main implication is false, then there is nothing to prove, because - again - the implication "false $rightarrow$ anything" is always true.
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
$endgroup$
add a comment |
$begingroup$
If the left-hand side of your main implication, i. e.
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$$
is true, then there are 2 possibilities:
Both $p$ and $r$ are true . Then both $q$ and $s$ must be true, too.
The right-hand side of your implication is then true, because it is reduced to "true $rightarrow$ true".
At least one of $p, r$ is false, i. e. $(p wedge r)$ is false. Then the right-hand side of your implication
$$left[ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
is true, because the implication "false $rightarrow$ anything" is always true.
If the left-hand side of your main implication is false, then there is nothing to prove, because - again - the implication "false $rightarrow$ anything" is always true.
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
$endgroup$
add a comment |
$begingroup$
If the left-hand side of your main implication, i. e.
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$$
is true, then there are 2 possibilities:
Both $p$ and $r$ are true . Then both $q$ and $s$ must be true, too.
The right-hand side of your implication is then true, because it is reduced to "true $rightarrow$ true".
At least one of $p, r$ is false, i. e. $(p wedge r)$ is false. Then the right-hand side of your implication
$$left[ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
is true, because the implication "false $rightarrow$ anything" is always true.
If the left-hand side of your main implication is false, then there is nothing to prove, because - again - the implication "false $rightarrow$ anything" is always true.
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
$endgroup$
If the left-hand side of your main implication, i. e.
$$left [ left ( prightarrow q right )wedge left ( rrightarrow s right ) right ]$$
is true, then there are 2 possibilities:
Both $p$ and $r$ are true . Then both $q$ and $s$ must be true, too.
The right-hand side of your implication is then true, because it is reduced to "true $rightarrow$ true".
At least one of $p, r$ is false, i. e. $(p wedge r)$ is false. Then the right-hand side of your implication
$$left[ left ( pwedge r right )rightarrow left ( qwedge s right ) right ]$$
is true, because the implication "false $rightarrow$ anything" is always true.
If the left-hand side of your main implication is false, then there is nothing to prove, because - again - the implication "false $rightarrow$ anything" is always true.
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
edited Mar 10 at 23:37
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
answered Mar 10 at 23:29
MarianDMarianD
1,2191614
1,2191614
add a comment |
add a comment |
1
$begingroup$
You want a proof using what axioms and rules of inference?
$endgroup$
– Ennar
Mar 10 at 22:50
$begingroup$
You have tagged the question using "boolean-algebra". What are the properties of the Boolean Algebra that you can use?
$endgroup$
– manooooh
Mar 10 at 22:51
$begingroup$
manoooooh(p→q means −p ∨ q that kinds of boolean algebra i think)
$endgroup$
– lucas
Mar 10 at 22:55
$begingroup$
ennar (I would not know how to apply rules of inference in this exercise.)
$endgroup$
– lucas
Mar 10 at 22:58
$begingroup$
$pto q$ is equivalent to $neg pvee q$ for any propositions $p$ and $q$. This law (called "Equivalence of conditional") can be used in any field of Logic, including Boolean Algebra.
$endgroup$
– manooooh
Mar 10 at 22:59