Mixing
Why $F rightarrow T implies neg F rightarrow neg T implies T rightarrow F$ is wrong? [closed]
$begingroup$
$$
F rightarrow T implies neg F rightarrow neg T implies T rightarrow F
$$
Why is this reasoning wrong according to the truth table of implication in logic?
Thanks.
discrete-mathematics logic
edited Jan 11 at 19:06
6005
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
$endgroup$
closed as off-topic by Adrian Keister, user91500, José Carlos Santos, amWhy, Song Jan 12 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, user91500, José Carlos Santos, amWhy, Song
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$
F rightarrow T implies neg F rightarrow neg T implies T rightarrow F
$$
Why is this reasoning wrong according to the truth table of implication in logic?
Thanks.
discrete-mathematics logic
edited Jan 11 at 19:06
6005
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
$endgroup$
closed as off-topic by Adrian Keister, user91500, José Carlos Santos, amWhy, Song Jan 12 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, user91500, José Carlos Santos, amWhy, Song
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48
add a comment |
$begingroup$
$$
F rightarrow T implies neg F rightarrow neg T implies T rightarrow F
$$
Why is this reasoning wrong according to the truth table of implication in logic?
Thanks.
discrete-mathematics logic
edited Jan 11 at 19:06
6005
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
$endgroup$
$$
F rightarrow T implies neg F rightarrow neg T implies T rightarrow F
$$
Why is this reasoning wrong according to the truth table of implication in logic?
Thanks.
discrete-mathematics logic
discrete-mathematics logic
edited Jan 11 at 19:06
6005
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
edited Jan 11 at 19:06
6005
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
edited Jan 11 at 19:06
6005
36.2k751125
edited Jan 11 at 19:06
6005
36.2k751125
edited Jan 11 at 19:06
6005
36.2k751125
36.2k751125
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
asked Jan 11 at 18:57
FriendlyFullStackFriendlyFullStack
62
62
closed as off-topic by Adrian Keister, user91500, José Carlos Santos, amWhy, Song Jan 12 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, user91500, José Carlos Santos, amWhy, Song
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Adrian Keister, user91500, José Carlos Santos, amWhy, Song Jan 12 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, user91500, José Carlos Santos, amWhy, Song
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48
add a comment |
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's go through your statements one at a time:
$$
F to T
$$
Looking at the truth table, this evaluates to $T$ (true).
$$
lnot F to lnot T
$$
(I use $lnot$ to denote "not") -- this one simplifies to $T to F$. Then looking at the truth table, it's $F$.
$$
T to F
$$
As we just saw, that is also $F$ (false).
So it looks like your problem is going from $F to T$ to $lnot F to lnot T$. That jump is not correct. In general,
$$
A to B text{ is equivalent to } lnot B to lnot A
$$
but
$$
A to B text{ is not necessarily equivalent to } lnot A to lnot B.
$$
answered Jan 11 at 19:03
60056005
36.2k751125
$endgroup$
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
add a comment |
$begingroup$
This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $rightarrow$ by performing the same operation on both sides.
I assume you started with
$$
F rightarrow T
$$
which is correct, and then proceeded to negate both sides like this
$$
neg F rightarrow neg T
$$
However this is not true because of how $rightarrow$ works with the implication.
answered Jan 11 at 19:06
wjmccannwjmccann
654118
$endgroup$
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's go through your statements one at a time:
$$
F to T
$$
Looking at the truth table, this evaluates to $T$ (true).
$$
lnot F to lnot T
$$
(I use $lnot$ to denote "not") -- this one simplifies to $T to F$. Then looking at the truth table, it's $F$.
$$
T to F
$$
As we just saw, that is also $F$ (false).
So it looks like your problem is going from $F to T$ to $lnot F to lnot T$. That jump is not correct. In general,
$$
A to B text{ is equivalent to } lnot B to lnot A
$$
but
$$
A to B text{ is not necessarily equivalent to } lnot A to lnot B.
$$
answered Jan 11 at 19:03
60056005
36.2k751125
$endgroup$
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
add a comment |
$begingroup$
Let's go through your statements one at a time:
$$
F to T
$$
Looking at the truth table, this evaluates to $T$ (true).
$$
lnot F to lnot T
$$
(I use $lnot$ to denote "not") -- this one simplifies to $T to F$. Then looking at the truth table, it's $F$.
$$
T to F
$$
As we just saw, that is also $F$ (false).
So it looks like your problem is going from $F to T$ to $lnot F to lnot T$. That jump is not correct. In general,
$$
A to B text{ is equivalent to } lnot B to lnot A
$$
but
$$
A to B text{ is not necessarily equivalent to } lnot A to lnot B.
$$
answered Jan 11 at 19:03
60056005
36.2k751125
$endgroup$
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
add a comment |
$begingroup$
Let's go through your statements one at a time:
$$
F to T
$$
Looking at the truth table, this evaluates to $T$ (true).
$$
lnot F to lnot T
$$
(I use $lnot$ to denote "not") -- this one simplifies to $T to F$. Then looking at the truth table, it's $F$.
$$
T to F
$$
As we just saw, that is also $F$ (false).
So it looks like your problem is going from $F to T$ to $lnot F to lnot T$. That jump is not correct. In general,
$$
A to B text{ is equivalent to } lnot B to lnot A
$$
but
$$
A to B text{ is not necessarily equivalent to } lnot A to lnot B.
$$
answered Jan 11 at 19:03
60056005
36.2k751125
$endgroup$
Let's go through your statements one at a time:
$$
F to T
$$
Looking at the truth table, this evaluates to $T$ (true).
$$
lnot F to lnot T
$$
(I use $lnot$ to denote "not") -- this one simplifies to $T to F$. Then looking at the truth table, it's $F$.
$$
T to F
$$
As we just saw, that is also $F$ (false).
So it looks like your problem is going from $F to T$ to $lnot F to lnot T$. That jump is not correct. In general,
$$
A to B text{ is equivalent to } lnot B to lnot A
$$
but
$$
A to B text{ is not necessarily equivalent to } lnot A to lnot B.
$$
answered Jan 11 at 19:03
60056005
36.2k751125
answered Jan 11 at 19:03
60056005
36.2k751125
answered Jan 11 at 19:03
60056005
36.2k751125
answered Jan 11 at 19:03
60056005
36.2k751125
36.2k751125
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
add a comment |
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
A→B is equivalent to ¬B→¬A this is what a was missing then. Thanks
$endgroup$
– FriendlyFullStack
Jan 11 at 19:09
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
@FriendlyFullStack You are welcome
$endgroup$
– 6005
Jan 11 at 19:12
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
$begingroup$
Could you please give me a reference containing the proof of this rule "A→B is equivalent to ¬B→¬A" ?
$endgroup$
– FriendlyFullStack
Jan 11 at 19:14
1
1
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
$begingroup$
For the curious ones: proof of rule "A→B is equivalent to ¬B→¬A" math.stackexchange.com/q/3070256/633862
$endgroup$
– FriendlyFullStack
Jan 11 at 19:44
add a comment |
$begingroup$
This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $rightarrow$ by performing the same operation on both sides.
I assume you started with
$$
F rightarrow T
$$
which is correct, and then proceeded to negate both sides like this
$$
neg F rightarrow neg T
$$
However this is not true because of how $rightarrow$ works with the implication.
answered Jan 11 at 19:06
wjmccannwjmccann
654118
$endgroup$
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
add a comment |
$begingroup$
This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $rightarrow$ by performing the same operation on both sides.
I assume you started with
$$
F rightarrow T
$$
which is correct, and then proceeded to negate both sides like this
$$
neg F rightarrow neg T
$$
However this is not true because of how $rightarrow$ works with the implication.
answered Jan 11 at 19:06
wjmccannwjmccann
654118
$endgroup$
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
add a comment |
$begingroup$
This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $rightarrow$ by performing the same operation on both sides.
I assume you started with
$$
F rightarrow T
$$
which is correct, and then proceeded to negate both sides like this
$$
neg F rightarrow neg T
$$
However this is not true because of how $rightarrow$ works with the implication.
answered Jan 11 at 19:06
wjmccannwjmccann
654118
$endgroup$
This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $rightarrow$ by performing the same operation on both sides.
I assume you started with
$$
F rightarrow T
$$
which is correct, and then proceeded to negate both sides like this
$$
neg F rightarrow neg T
$$
However this is not true because of how $rightarrow$ works with the implication.
answered Jan 11 at 19:06
wjmccannwjmccann
654118
edited Jan 11 at 19:12
answered Jan 11 at 19:06
wjmccannwjmccann
654118
answered Jan 11 at 19:06
wjmccannwjmccann
654118
answered Jan 11 at 19:06
wjmccannwjmccann
654118
654118
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
add a comment |
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
1
1
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
" unlike algebra, you do not preserve truth in logic by performing the same operation on both sides" I think that's misleading. The point isn't that logic is different from algebra, it's that "$rightarrow$" is different from "$=$." E.g. you can't do the same thing to both sides of an inequality in algebra, either: $2<3notrightarrow -2<-3$.
$endgroup$
– Noah Schweber
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber Beat me to it.
$endgroup$
– J.G.
Jan 11 at 19:09
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
$begingroup$
@NoahSchweber tried to make it more clear what I was saying
$endgroup$
– wjmccann
Jan 11 at 19:13
add a comment |
$begingroup$
Focus on $F to T implieslnot F tolnot T$, as $lnot F tolnot T implies T to F$ is correct.
$endgroup$
– Arthur
Jan 11 at 19:01
$begingroup$
I don't understand what you're asking. Are $T$ and $F$ truth values? Or are they statements? Also, what is the distinction you're making between "$rightarrow$" and "$Rightarrow$"?
$endgroup$
– Joe
Jan 11 at 19:01
$begingroup$
What is $implies$ -- does that mean "simplify to"?
$endgroup$
– 6005
Jan 11 at 19:02
$begingroup$
I'd argue that F -> T implies not T -> not F
$endgroup$
– Andreas
Jan 11 at 19:03
$begingroup$
It's a tautology if $T$ and $F$ are logical propositions (unusual) and $implies$ is equivalent to $to$.
$endgroup$
– Dan Christensen
Jan 12 at 17:48