Mixing
Show (p ∧ q) → q ≡ T by using table 6 and the first line of table 7
$begingroup$
I used the left side to attempt to prove it was T.
Chart of Laws
(p ∧ q) → q ≡ T
(p ∧ q) → q
Using the first line in table 7
¬(p ∧ q) v q
D'Morgans Law
(¬p v ¬q) v q
Associative Law
¬p v (¬q v q)
Negation Law
¬p v T
Domination Law
T ≡ T
I was wondering if this is right since I felt that you can't do that with domination law. If anyone could double check thank you.
discrete-mathematics logic
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
$endgroup$
add a comment |
$begingroup$
I used the left side to attempt to prove it was T.
Chart of Laws
(p ∧ q) → q ≡ T
(p ∧ q) → q
Using the first line in table 7
¬(p ∧ q) v q
D'Morgans Law
(¬p v ¬q) v q
Associative Law
¬p v (¬q v q)
Negation Law
¬p v T
Domination Law
T ≡ T
I was wondering if this is right since I felt that you can't do that with domination law. If anyone could double check thank you.
discrete-mathematics logic
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
$endgroup$
$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
1
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43
add a comment |
$begingroup$
I used the left side to attempt to prove it was T.
Chart of Laws
(p ∧ q) → q ≡ T
(p ∧ q) → q
Using the first line in table 7
¬(p ∧ q) v q
D'Morgans Law
(¬p v ¬q) v q
Associative Law
¬p v (¬q v q)
Negation Law
¬p v T
Domination Law
T ≡ T
I was wondering if this is right since I felt that you can't do that with domination law. If anyone could double check thank you.
discrete-mathematics logic
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
$endgroup$
I used the left side to attempt to prove it was T.
Chart of Laws
(p ∧ q) → q ≡ T
(p ∧ q) → q
Using the first line in table 7
¬(p ∧ q) v q
D'Morgans Law
(¬p v ¬q) v q
Associative Law
¬p v (¬q v q)
Negation Law
¬p v T
Domination Law
T ≡ T
I was wondering if this is right since I felt that you can't do that with domination law. If anyone could double check thank you.
discrete-mathematics logic
discrete-mathematics logic
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
edited Jan 25 at 0:26
hentaidaddy666
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
asked Jan 25 at 0:17
hentaidaddy666hentaidaddy666
112
112
$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
1
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43
add a comment |
$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
1
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43
$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
1
1
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes that's completely correct.
The domination law states that $q lor T$ is always true for any statement $q$. So it's true especially for any negated statement $neg p equiv q$.
Another way to think about this is to consider:
$$(neg p lor T) equiv negneg (neg p lor T) equiv neg(p land F) equiv neg F equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $land$ operator.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes that's completely correct.
The domination law states that $q lor T$ is always true for any statement $q$. So it's true especially for any negated statement $neg p equiv q$.
Another way to think about this is to consider:
$$(neg p lor T) equiv negneg (neg p lor T) equiv neg(p land F) equiv neg F equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $land$ operator.
$endgroup$
add a comment |
$begingroup$
Yes that's completely correct.
The domination law states that $q lor T$ is always true for any statement $q$. So it's true especially for any negated statement $neg p equiv q$.
Another way to think about this is to consider:
$$(neg p lor T) equiv negneg (neg p lor T) equiv neg(p land F) equiv neg F equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $land$ operator.
$endgroup$
add a comment |
$begingroup$
Yes that's completely correct.
The domination law states that $q lor T$ is always true for any statement $q$. So it's true especially for any negated statement $neg p equiv q$.
Another way to think about this is to consider:
$$(neg p lor T) equiv negneg (neg p lor T) equiv neg(p land F) equiv neg F equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $land$ operator.
$endgroup$
Yes that's completely correct.
The domination law states that $q lor T$ is always true for any statement $q$. So it's true especially for any negated statement $neg p equiv q$.
Another way to think about this is to consider:
$$(neg p lor T) equiv negneg (neg p lor T) equiv neg(p land F) equiv neg F equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $land$ operator.
edited Jan 25 at 3:16
answered Jan 25 at 0:24
user635162
add a comment |
add a comment |
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$begingroup$
Yes, that's fine ... the $p$'s, $q$'s and $r$'s in these laws can be any statement
$endgroup$
– Bram28
Jan 25 at 0:42
1
$begingroup$
Only one little error, it's De Morgan, not D'Morgan
$endgroup$
– Bernard Massé
Jan 25 at 0:43