Mixing
The meaning of implication in logic
How to remember implication logic by remembering a simple english.
I read some sentence like
if P,then Q
P only if Q
Q if P
But i am unable to correlate these sentences with the following logic. Although truth table is very simple but i don't want to just remember it without it's actual meaning.
P Q P=>Q
0 0 1
0 1 1
1 0 0
1 1 1
logic
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
asked Jan 18 '12 at 23:04
P K
506259
add a comment |
How to remember implication logic by remembering a simple english.
I read some sentence like
if P,then Q
P only if Q
Q if P
But i am unable to correlate these sentences with the following logic. Although truth table is very simple but i don't want to just remember it without it's actual meaning.
P Q P=>Q
0 0 1
0 1 1
1 0 0
1 1 1
logic
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
asked Jan 18 '12 at 23:04
P K
506259
5
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45
add a comment |
How to remember implication logic by remembering a simple english.
I read some sentence like
if P,then Q
P only if Q
Q if P
But i am unable to correlate these sentences with the following logic. Although truth table is very simple but i don't want to just remember it without it's actual meaning.
P Q P=>Q
0 0 1
0 1 1
1 0 0
1 1 1
logic
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
asked Jan 18 '12 at 23:04
P K
506259
How to remember implication logic by remembering a simple english.
I read some sentence like
if P,then Q
P only if Q
Q if P
But i am unable to correlate these sentences with the following logic. Although truth table is very simple but i don't want to just remember it without it's actual meaning.
P Q P=>Q
0 0 1
0 1 1
1 0 0
1 1 1
logic
logic
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
asked Jan 18 '12 at 23:04
P K
506259
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
asked Jan 18 '12 at 23:04
P K
506259
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
edited Jan 18 '12 at 23:14
Austin Mohr
20k35097
20k35097
asked Jan 18 '12 at 23:04
P K
506259
asked Jan 18 '12 at 23:04
P K
506259
asked Jan 18 '12 at 23:04
P K
506259
506259
5
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45
add a comment |
5
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45
5
5
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45
add a comment |
9 Answers
9
active
oldest
votes
If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)
If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
add a comment |
Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).
A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).
You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
add a comment |
Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.
answered Jan 20 '12 at 5:12
plutoniumium
413
add a comment |
"if P then Q" is equivalent to P=>Q
"P only if Q" is also equivalent to P=>Q (example here)
"Q if P" is same as "if P then Q" and equivalent to P=>Q
However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
add a comment |
In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
add a comment |
Here's a good rule to follow when you're at a university (or working on the job):
"If the fire alarm is going off, you are to go outside"
This can be expressed as P -> Q:
P = "Is the alarm going off?"
Q = "Are you going outside?"
begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}
answered Nov 20 '18 at 19:31
DavidHulsman
111
add a comment |
Here's your truth table:
$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$
$1$ means true and $0$ means false.
What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
add a comment |
From the MIT CS Math course
The truth table for implications can be summarized in words as
follows:
An implication is true exactly when the if-part is false or the
then-part is true.
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
add a comment |
I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.
Example:
You have a nice kid and you make him the following promise to him:
If you get an A in your exam, then I will buy you a car.
In this case P is kid gets A in exam and Q is You buy him a car.
Now let's see how this promise holds with various values for P and Q
If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.
If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.
If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.
If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
add a comment |
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9 Answers
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9 Answers
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If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)
If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
add a comment |
If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)
If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
add a comment |
If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)
If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
If you start out with a false premise, then, as far as implication is concerned, you are free to conclude anything. (This corresponds to the fact that, when $P$ is false, the implication $P rightarrow Q$ is true no matter what $Q$ is.)
If you start out with a true premise, then the implication should be true only when the conclusion is also true. (This corresponds to the fact that, when $P$ is true, the truth of the implication is the same as the truth of $Q$.)
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
answered Jan 18 '12 at 23:11
Austin Mohr
20k35097
20k35097
add a comment |
add a comment |
Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).
A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).
You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
add a comment |
Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).
A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).
You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
add a comment |
Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).
A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).
You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
Not really as an answer but as an anecdote I'll sketch the following real life situation (in abstract terms to avoid controversy).
A politician $s$ declares, in a menacing voice: "if we would do $P$ then $Q$ will ensue!" where $P$ is something like electing his adversary, or not adopting the Draconic measures he proposes, and $Q$ are catastrophic events like people losing their jobs and the country plunging into a deep crisis. Now suppose $s$ is lucky and manages to avoid $P$, but that then $Q$ happens anyway. Now does this show that $s$ lied? Since $P$ is false and $Q$ is true, we are in the second line of your table, you can read off that $PRightarrow Q$ is deemed true in this case. In fact since $s$ prophesized about a circumstance $P$ that did not happen, later events could not have shown him a liar either way. An this in spite of the fact that by common sense the statement he made was either false (if doing $P$ would actually have prevented $Q$) or irrelevant (if $Q$ would have happened independently of $P$, or depending on other conditions than those of $P$).
You see how smart politicians are? (Of course $s$ can be shown to be a liar if he does not manage to avoid $P$, but then being out of office anyway, $s$ probably won't care much about being proven a liar as well.)
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
answered Jan 19 '12 at 7:41
Marc van Leeuwen
86.4k5106219
86.4k5106219
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
add a comment |
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
This anecdote leads to the answer; or at least the explanation of why the truth table for "If P then Q" doesn't quite match what some people expect. In everyday situations, when we say "If ... then ...", this actually implies some degree of causation; that is, that P somehow causes Q, or makes Q more likely. Of course, propositional logic doesn't express causation or likelihood; so the propositional logic meaning of "If P then Q" differs ever-so-slightly from the everyday meaning.
– user22805
Jan 19 '12 at 9:21
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
Thanks Marc for nice explanation.
– P K
Jan 19 '12 at 17:54
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
thanks Marc! This really helped me understand implication :)
– ambertch
Aug 5 '12 at 23:25
add a comment |
Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.
answered Jan 20 '12 at 5:12
plutoniumium
413
add a comment |
Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.
answered Jan 20 '12 at 5:12
plutoniumium
413
add a comment |
Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.
answered Jan 20 '12 at 5:12
plutoniumium
413
Remember that "implies" is equivalent to "subset of". It works in exactly the same way: "if an element is in the subset (e.g A), it MUST also be in the superset (e.g. B)". By definition, it is impossible that an element is in the subset, but not in the superset. That's the P=1, Q=0; P=>Q = 0 case. In fact, "A ⊆ B" means that a ∈ A implies that a ∈ B. If a is not in subset A then you can't draw any conclusions on whether a is in the superset B. That's how I keep remembering it.
answered Jan 20 '12 at 5:12
plutoniumium
413
answered Jan 20 '12 at 5:12
plutoniumium
413
answered Jan 20 '12 at 5:12
plutoniumium
413
answered Jan 20 '12 at 5:12
plutoniumium
413
413
add a comment |
add a comment |
"if P then Q" is equivalent to P=>Q
"P only if Q" is also equivalent to P=>Q (example here)
"Q if P" is same as "if P then Q" and equivalent to P=>Q
However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
add a comment |
"if P then Q" is equivalent to P=>Q
"P only if Q" is also equivalent to P=>Q (example here)
"Q if P" is same as "if P then Q" and equivalent to P=>Q
However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
add a comment |
"if P then Q" is equivalent to P=>Q
"P only if Q" is also equivalent to P=>Q (example here)
"Q if P" is same as "if P then Q" and equivalent to P=>Q
However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
"if P then Q" is equivalent to P=>Q
"P only if Q" is also equivalent to P=>Q (example here)
"Q if P" is same as "if P then Q" and equivalent to P=>Q
However, note (following statement which is not given in the original question): P if Q is equivalent to "if q then p" or Q=>P
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
answered Jan 18 '12 at 23:18
Sniper Clown
51521333
51521333
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
add a comment |
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
linked example makes sense to me. thanks.
– P K
Jan 18 '12 at 23:25
add a comment |
In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
add a comment |
In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
add a comment |
In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
In your truth table, look only at the lines where $Pimplies Q$ holds (is $1$), i.e., drop the third line. In the remaining lines, for each line where $P$ holds (i.e., the last one) $Q$ holds as well. Moreover, the column $Pimplies Q$ has $1$s in all lines possible where this property holds (i.e., if we made its unique entry $0$ in the third line a $1$ as well, the property would fail).
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
answered Jan 18 '12 at 23:13
Marc van Leeuwen
86.4k5106219
86.4k5106219
add a comment |
add a comment |
Here's a good rule to follow when you're at a university (or working on the job):
"If the fire alarm is going off, you are to go outside"
This can be expressed as P -> Q:
P = "Is the alarm going off?"
Q = "Are you going outside?"
begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}
answered Nov 20 '18 at 19:31
DavidHulsman
111
add a comment |
Here's a good rule to follow when you're at a university (or working on the job):
"If the fire alarm is going off, you are to go outside"
This can be expressed as P -> Q:
P = "Is the alarm going off?"
Q = "Are you going outside?"
begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}
answered Nov 20 '18 at 19:31
DavidHulsman
111
add a comment |
Here's a good rule to follow when you're at a university (or working on the job):
"If the fire alarm is going off, you are to go outside"
This can be expressed as P -> Q:
P = "Is the alarm going off?"
Q = "Are you going outside?"
begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}
answered Nov 20 '18 at 19:31
DavidHulsman
111
Here's a good rule to follow when you're at a university (or working on the job):
"If the fire alarm is going off, you are to go outside"
This can be expressed as P -> Q:
P = "Is the alarm going off?"
Q = "Are you going outside?"
begin{array}{|r|r|r|} text{P} & text{Q} & text{Answer} & text{Have you followed the rule?} \ hline
text{F} & text{F} & text{T} & text{Yes, the rule has been followed} \ hline
text{F} & text{T} & text{T} & text{Yes, the rule has been followed (you can still go outside during a break)} \ hline
text{T} & text{F} & text{F} & text{No, the rule has been broken!} \ hline
text{T} & text{T} & text{T} & text{Yes, the rule has been followed} \ hline
hline
end{array}
answered Nov 20 '18 at 19:31
DavidHulsman
111
answered Nov 20 '18 at 19:31
DavidHulsman
111
answered Nov 20 '18 at 19:31
DavidHulsman
111
answered Nov 20 '18 at 19:31
DavidHulsman
111
111
add a comment |
add a comment |
Here's your truth table:
$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$
$1$ means true and $0$ means false.
What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
add a comment |
Here's your truth table:
$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$
$1$ means true and $0$ means false.
What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
add a comment |
Here's your truth table:
$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$
$1$ means true and $0$ means false.
What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
Here's your truth table:
$$ begin{align}mathbf P & mathbf Q & mathbf{P implies Q} \ 1 & 1&1 \ 1&0&0 \ 0&1&1 \ 0&0&1end{align}$$
$1$ means true and $0$ means false.
What does logical implication mean? "If $Phi$ then $Psi$" can be written as $mathbf{Phi Rightarrow Psi}$. Albeit, it's much more defined than real life. Remember that Mathematical thinking is different than general thinking.
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
answered Sep 25 '12 at 15:29
Parth Kohli
6,01012860
6,01012860
add a comment |
add a comment |
From the MIT CS Math course
The truth table for implications can be summarized in words as
follows:
An implication is true exactly when the if-part is false or the
then-part is true.
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
add a comment |
From the MIT CS Math course
The truth table for implications can be summarized in words as
follows:
An implication is true exactly when the if-part is false or the
then-part is true.
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
add a comment |
From the MIT CS Math course
The truth table for implications can be summarized in words as
follows:
An implication is true exactly when the if-part is false or the
then-part is true.
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
From the MIT CS Math course
The truth table for implications can be summarized in words as
follows:
An implication is true exactly when the if-part is false or the
then-part is true.
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
answered Jun 18 '14 at 6:20
Ahmad Ajmi
1011
1011
add a comment |
add a comment |
I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.
Example:
You have a nice kid and you make him the following promise to him:
If you get an A in your exam, then I will buy you a car.
In this case P is kid gets A in exam and Q is You buy him a car.
Now let's see how this promise holds with various values for P and Q
If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.
If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.
If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.
If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
add a comment |
I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.
Example:
You have a nice kid and you make him the following promise to him:
If you get an A in your exam, then I will buy you a car.
In this case P is kid gets A in exam and Q is You buy him a car.
Now let's see how this promise holds with various values for P and Q
If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.
If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.
If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.
If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
add a comment |
I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.
Example:
You have a nice kid and you make him the following promise to him:
If you get an A in your exam, then I will buy you a car.
In this case P is kid gets A in exam and Q is You buy him a car.
Now let's see how this promise holds with various values for P and Q
If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.
If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.
If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.
If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
I would like to share my own understanding of this.
I like to think of Implication as a Promise rather than Causality which is the natural tendency when you come across it the first time.
Example:
You have a nice kid and you make him the following promise to him:
If you get an A in your exam, then I will buy you a car.
In this case P is kid gets A in exam and Q is You buy him a car.
Now let's see how this promise holds with various values for P and Q
If P is true (Kid gets A in exam) and Q is true (You bought him car) then your promise has held and $P Rightarrow Q$ is true.
If P is true (Kid gets A in exam) and Q is false (You didn't buy him a car) then your promise didn't hold so $P Rightarrow Q$ is false.
If P is false (Kid didn't get A in exam) and Q is true (You bought him car) then your promise still holds and $P Rightarrow Q$ is true and that's because you only said what will happen if he get's an A, you basically didn't say what will happen if he doesn't which could imply anything. Basically you didn't break your promise and this is the weak property which most people find confusing in implication.
If P is false (Kid didn't get A in exam) and Q is false (you didn't buy him a car) then your promise has also held and $P Rightarrow Q$ is true because you only promised and guaranteed a car if he gets an A.
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
answered Oct 22 '17 at 19:05
Ibrahim Najjar
1385
1385
add a comment |
add a comment |
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5
It may help to read the implication "$P implies Q$" as "buy $P$, get $Q$ free (whether you want it or not!)". Buying $P$ means that you also get $Q$, so you have both; not-buying $P$ doesn't rule out the possibility of getting (or not-getting) $Q$ by other means; however, not-getting $Q$ does rule out having bought $P$.
– Blue
Jan 20 '12 at 6:04
See also math.stackexchange.com/questions/439987/…
– amWhy
Jan 10 '18 at 20:45