The Logic of False Statements [closed]
I"m wondering whether it's valid to carry logical operations across false statements, same as as you would across true statements.
Compare:
$$begin{alignat*}{3}
& 1 < 2 \
Rightarrow & 1 + 1 < 2 + 1 \
Rightarrow & frac{1}{3} < frac{1}{2}\
end{alignat*}$$
With:
$$begin{alignat*}{3}
& 2 < 1 &&hspace{2cm} text{is false} \
Rightarrow & 2 + 1 < 1 + 1 &&hspace{2cm} text{is false} \
Rightarrow & frac{1}{2} < frac {1}{3} &&hspace{2cm} text{is false} \
end{alignat*}$$
logic propositional-calculus
closed as unclear what you're asking by amWhy, Cesareo, Lord Shark the Unknown, max_zorn, Gibbs Nov 27 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I"m wondering whether it's valid to carry logical operations across false statements, same as as you would across true statements.
Compare:
$$begin{alignat*}{3}
& 1 < 2 \
Rightarrow & 1 + 1 < 2 + 1 \
Rightarrow & frac{1}{3} < frac{1}{2}\
end{alignat*}$$
With:
$$begin{alignat*}{3}
& 2 < 1 &&hspace{2cm} text{is false} \
Rightarrow & 2 + 1 < 1 + 1 &&hspace{2cm} text{is false} \
Rightarrow & frac{1}{2} < frac {1}{3} &&hspace{2cm} text{is false} \
end{alignat*}$$
logic propositional-calculus
closed as unclear what you're asking by amWhy, Cesareo, Lord Shark the Unknown, max_zorn, Gibbs Nov 27 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29
add a comment |
I"m wondering whether it's valid to carry logical operations across false statements, same as as you would across true statements.
Compare:
$$begin{alignat*}{3}
& 1 < 2 \
Rightarrow & 1 + 1 < 2 + 1 \
Rightarrow & frac{1}{3} < frac{1}{2}\
end{alignat*}$$
With:
$$begin{alignat*}{3}
& 2 < 1 &&hspace{2cm} text{is false} \
Rightarrow & 2 + 1 < 1 + 1 &&hspace{2cm} text{is false} \
Rightarrow & frac{1}{2} < frac {1}{3} &&hspace{2cm} text{is false} \
end{alignat*}$$
logic propositional-calculus
I"m wondering whether it's valid to carry logical operations across false statements, same as as you would across true statements.
Compare:
$$begin{alignat*}{3}
& 1 < 2 \
Rightarrow & 1 + 1 < 2 + 1 \
Rightarrow & frac{1}{3} < frac{1}{2}\
end{alignat*}$$
With:
$$begin{alignat*}{3}
& 2 < 1 &&hspace{2cm} text{is false} \
Rightarrow & 2 + 1 < 1 + 1 &&hspace{2cm} text{is false} \
Rightarrow & frac{1}{2} < frac {1}{3} &&hspace{2cm} text{is false} \
end{alignat*}$$
logic propositional-calculus
logic propositional-calculus
edited Nov 24 at 16:05
asked Nov 24 at 15:24
Miguel Cumming-Romo
155
155
closed as unclear what you're asking by amWhy, Cesareo, Lord Shark the Unknown, max_zorn, Gibbs Nov 27 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Cesareo, Lord Shark the Unknown, max_zorn, Gibbs Nov 27 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29
add a comment |
1
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29
1
1
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29
add a comment |
4 Answers
4
active
oldest
votes
$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).
Logical operations are carried on statements, either true or false.
But valid logical operations license the truth of the conclusion only when applied to true premises.
add a comment |
If I understand you right, you're asking, in essence,
If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?
No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.
The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.
Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.
add a comment |
Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.
Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
add a comment |
No.
2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).
Logical operations are carried on statements, either true or false.
But valid logical operations license the truth of the conclusion only when applied to true premises.
add a comment |
$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).
Logical operations are carried on statements, either true or false.
But valid logical operations license the truth of the conclusion only when applied to true premises.
add a comment |
$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).
Logical operations are carried on statements, either true or false.
But valid logical operations license the truth of the conclusion only when applied to true premises.
$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).
Logical operations are carried on statements, either true or false.
But valid logical operations license the truth of the conclusion only when applied to true premises.
answered Nov 24 at 15:49
Mauro ALLEGRANZA
64.1k448111
64.1k448111
add a comment |
add a comment |
If I understand you right, you're asking, in essence,
If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?
No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.
The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.
Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.
add a comment |
If I understand you right, you're asking, in essence,
If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?
No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.
The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.
Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.
add a comment |
If I understand you right, you're asking, in essence,
If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?
No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.
The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.
Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.
If I understand you right, you're asking, in essence,
If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?
No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.
The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.
Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.
answered Nov 25 at 14:45
Henning Makholm
236k16300535
236k16300535
add a comment |
add a comment |
Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.
Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
add a comment |
Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.
Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
add a comment |
Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.
Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$
Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.
Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$
answered Nov 24 at 16:02
Bram28
59.6k44186
59.6k44186
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
add a comment |
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
Modus ponens only works when the antecedent is true and the conditional is true. If the antecedent is false,and the conditional is true, it isn't valid to infer that the consequent is false.
– Doug Spoonwood
Nov 25 at 12:44
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
I wasn’t saying that since the antecedent is false then the consequent is false... i was just showing a case where you can make a valid inference on false statements, which I thought the question was about ... but maybe I misunderstood the question.
– Bram28
Nov 25 at 12:50
add a comment |
No.
2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.
add a comment |
No.
2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.
add a comment |
No.
2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.
No.
2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.
answered Nov 25 at 12:48
Doug Spoonwood
8,00112144
8,00112144
add a comment |
add a comment |
1
Not clear... From $1 < 2$ you add $1$ to both to get correctly $2 < 3$. Then how you derive ; $dfrac 1 2 < 1$ ?
– Mauro ALLEGRANZA
Nov 24 at 15:29