Is it sufficient to prove $P(x) geq a$ if we already know $P(x) > a$?
$begingroup$
Is it sufficient to prove $P(x) ge (text{or} le) a$ if we already know $P(x) > (text{or} <)a$?
For example, to prove
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} le 2
$$
Suppose I have already proved
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} < frac{7}{4} - frac{1}{n}
$$
Then, are we done? (Because $frac{7}{4}-frac{1}{n} < 2 le 2$)
discrete-mathematics theorem-provers
edited Jan 12 at 9:10
rtybase
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
$endgroup$
add a comment |
$begingroup$
Is it sufficient to prove $P(x) ge (text{or} le) a$ if we already know $P(x) > (text{or} <)a$?
For example, to prove
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} le 2
$$
Suppose I have already proved
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} < frac{7}{4} - frac{1}{n}
$$
Then, are we done? (Because $frac{7}{4}-frac{1}{n} < 2 le 2$)
discrete-mathematics theorem-provers
edited Jan 12 at 9:10
rtybase
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
$endgroup$
add a comment |
$begingroup$
Is it sufficient to prove $P(x) ge (text{or} le) a$ if we already know $P(x) > (text{or} <)a$?
For example, to prove
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} le 2
$$
Suppose I have already proved
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} < frac{7}{4} - frac{1}{n}
$$
Then, are we done? (Because $frac{7}{4}-frac{1}{n} < 2 le 2$)
discrete-mathematics theorem-provers
edited Jan 12 at 9:10
rtybase
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
$endgroup$
Is it sufficient to prove $P(x) ge (text{or} le) a$ if we already know $P(x) > (text{or} <)a$?
For example, to prove
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} le 2
$$
Suppose I have already proved
$$
forall n ge 1 , sum_{i=1}^{n}frac{1}{i^2} < frac{7}{4} - frac{1}{n}
$$
Then, are we done? (Because $frac{7}{4}-frac{1}{n} < 2 le 2$)
discrete-mathematics theorem-provers
discrete-mathematics theorem-provers
edited Jan 12 at 9:10
rtybase
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
edited Jan 12 at 9:10
rtybase
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
edited Jan 12 at 9:10
rtybase
11.4k31533
edited Jan 12 at 9:10
rtybase
11.4k31533
edited Jan 12 at 9:10
rtybase
11.4k31533
11.4k31533
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
asked Jan 12 at 6:26
ShinobuIsMyWifeShinobuIsMyWife
463
463
add a comment |
add a comment |
1 Answer
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$begingroup$
Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
add a comment |
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1 Answer
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active
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1 Answer
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votes
$begingroup$
Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
add a comment |
$begingroup$
Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
add a comment |
$begingroup$
Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 12 at 6:48
PiKindOfGuyPiKindOfGuy
18611
18611
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
add a comment |
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
$begingroup$
Opps! I just found that $le$ means less that or equal to in English, so it is actually kind of a logic union.
$endgroup$
– ShinobuIsMyWife
Jan 12 at 6:53
add a comment |
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