Mixing
What is the negation of $pto sim q$?
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I know that the negation of $pto q$ is ~p V q but I can’t seem to figure out the effect $sim q$ will have on the negation. Also is their a way to check if something is the negation of a statement?
propositional-calculus
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
$endgroup$
add a comment |
$begingroup$
I know that the negation of $pto q$ is ~p V q but I can’t seem to figure out the effect $sim q$ will have on the negation. Also is their a way to check if something is the negation of a statement?
propositional-calculus
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
$endgroup$
1
$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
2
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09
add a comment |
$begingroup$
I know that the negation of $pto q$ is ~p V q but I can’t seem to figure out the effect $sim q$ will have on the negation. Also is their a way to check if something is the negation of a statement?
propositional-calculus
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
$endgroup$
I know that the negation of $pto q$ is ~p V q but I can’t seem to figure out the effect $sim q$ will have on the negation. Also is their a way to check if something is the negation of a statement?
propositional-calculus
propositional-calculus
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
edited Jan 23 at 7:23
Mauro ALLEGRANZA
67k449115
67k449115
asked Jan 22 at 21:53
MikaylaMikayla
114
asked Jan 22 at 21:53
MikaylaMikayla
114
asked Jan 22 at 21:53
MikaylaMikayla
114
114
1
$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
2
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09
add a comment |
1
$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
2
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09
1
1
$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
2
2
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The definition of implication in propositional logic can be expressed with
begin{equation}tag{1}
prightarrow q~:=~(sim p)lor q
end{equation}
By this definition, the negation of $prightarrow q$ is $pland(sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $prightarrow (sim q)$ is
begin{align}
sim(prightarrow (sim q))~~~&Start\
sim((sim p)lor (sim q))~~~&Substitute~(1)\
pland q~~~&De~Morgan's~law
end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
answered Jan 22 at 23:35
user400188user400188
4241614
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add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
The definition of implication in propositional logic can be expressed with
begin{equation}tag{1}
prightarrow q~:=~(sim p)lor q
end{equation}
By this definition, the negation of $prightarrow q$ is $pland(sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $prightarrow (sim q)$ is
begin{align}
sim(prightarrow (sim q))~~~&Start\
sim((sim p)lor (sim q))~~~&Substitute~(1)\
pland q~~~&De~Morgan's~law
end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
answered Jan 22 at 23:35
user400188user400188
4241614
$endgroup$
add a comment |
$begingroup$
The definition of implication in propositional logic can be expressed with
begin{equation}tag{1}
prightarrow q~:=~(sim p)lor q
end{equation}
By this definition, the negation of $prightarrow q$ is $pland(sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $prightarrow (sim q)$ is
begin{align}
sim(prightarrow (sim q))~~~&Start\
sim((sim p)lor (sim q))~~~&Substitute~(1)\
pland q~~~&De~Morgan's~law
end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
answered Jan 22 at 23:35
user400188user400188
4241614
$endgroup$
add a comment |
$begingroup$
The definition of implication in propositional logic can be expressed with
begin{equation}tag{1}
prightarrow q~:=~(sim p)lor q
end{equation}
By this definition, the negation of $prightarrow q$ is $pland(sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $prightarrow (sim q)$ is
begin{align}
sim(prightarrow (sim q))~~~&Start\
sim((sim p)lor (sim q))~~~&Substitute~(1)\
pland q~~~&De~Morgan's~law
end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
answered Jan 22 at 23:35
user400188user400188
4241614
$endgroup$
The definition of implication in propositional logic can be expressed with
begin{equation}tag{1}
prightarrow q~:=~(sim p)lor q
end{equation}
By this definition, the negation of $prightarrow q$ is $pland(sim q)$. This means that the statement $p$ implies $q$, is false only when the antecedent or premise of the statement is true, and its consequent false.
Using the same definition, negation of the statement $prightarrow (sim q)$ is
begin{align}
sim(prightarrow (sim q))~~~&Start\
sim((sim p)lor (sim q))~~~&Substitute~(1)\
pland q~~~&De~Morgan's~law
end{align}
To check if any two statements in propositional logic are the negation of one another, it is sufficient to substitute all instances of implication with the first definition, then negate one, and apply De Morgan's law successively to it until one reduces to the other. If this is the case, then they are negations of one-another.
answered Jan 22 at 23:35
user400188user400188
4241614
answered Jan 22 at 23:35
user400188user400188
4241614
answered Jan 22 at 23:35
user400188user400188
4241614
answered Jan 22 at 23:35
user400188user400188
4241614
4241614
add a comment |
add a comment |
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$begingroup$
Welcome to Maths SX! I'm sorry, but the negation of $pto q$ is $pwedge ({sim}q)$ since $pto q$ is $({sim}p)vee q$.
$endgroup$
– Bernard
Jan 22 at 21:57
2
$begingroup$
The negation for $pto q$ is $plandlnot q$. So by substitution ...
$endgroup$
– Graham Kemp
Jan 22 at 21:59
$begingroup$
Caution. The negation of $pto q$ is ${sim (p}to q)$ but it is not $({sim p})to q$. Which did you mean when you wrote ${sim p}to q$? So in the same way, the negation of $p to sim q$ is $sim (p to sim q)$.
$endgroup$
– GEdgar
Jan 22 at 22:44
$begingroup$
Whoops meant to write the negation as ~p V q
$endgroup$
– Mikayla
Jan 22 at 23:09