Mixing
Negation of “and” statements: a and b
$begingroup$
Is it correct?
$neg$(a and b)=(not a) or (not b)
What ruleset can i look up for negations? Especially for "all", "if, then" statements.
logic
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
$endgroup$
add a comment |
$begingroup$
Is it correct?
$neg$(a and b)=(not a) or (not b)
What ruleset can i look up for negations? Especially for "all", "if, then" statements.
logic
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
$endgroup$
2
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02
add a comment |
$begingroup$
Is it correct?
$neg$(a and b)=(not a) or (not b)
What ruleset can i look up for negations? Especially for "all", "if, then" statements.
logic
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
$endgroup$
Is it correct?
$neg$(a and b)=(not a) or (not b)
What ruleset can i look up for negations? Especially for "all", "if, then" statements.
logic
logic
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
asked Oct 23 '16 at 0:51
AlucardAlucard
8751627
8751627
2
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02
add a comment |
2
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02
2
2
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02
add a comment |
1 Answer
1
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$begingroup$
Yes, that's called De Morgan's Laws. This site has more rules about negations of logical connectives and this PDF should help you with negation of universal and existential quantifiers.
edited Jan 24 at 22:57
Community♦
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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$begingroup$
Yes, that's called De Morgan's Laws. This site has more rules about negations of logical connectives and this PDF should help you with negation of universal and existential quantifiers.
edited Jan 24 at 22:57
Community♦
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Yes, that's called De Morgan's Laws. This site has more rules about negations of logical connectives and this PDF should help you with negation of universal and existential quantifiers.
edited Jan 24 at 22:57
Community♦
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Yes, that's called De Morgan's Laws. This site has more rules about negations of logical connectives and this PDF should help you with negation of universal and existential quantifiers.
edited Jan 24 at 22:57
Community♦
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
Yes, that's called De Morgan's Laws. This site has more rules about negations of logical connectives and this PDF should help you with negation of universal and existential quantifiers.
edited Jan 24 at 22:57
Community♦
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
edited Jan 24 at 22:57
Community♦
1
edited Jan 24 at 22:57
Community♦
1
edited Jan 24 at 22:57
Community♦
1
1
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
answered Oct 23 '16 at 1:12
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
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2
$begingroup$
That is correct. Quatifiers, when negated change to the other, ex $forall$ becomes $exists$. If P, the Q is equivalent to $lnot$ P $lor$ Q. You can use the negation of that to obtain the negation of the implication (if, then).
$endgroup$
– RJM
Oct 23 '16 at 1:02