Mixing
Strong induction and vacuous truth
$begingroup$
I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states:
Strong induction proves a sequence of statements $P(0), P(1), …$ by
proving the implication
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$,
then $P(n)$ is true."
for every nonnegative integer $n$. There is no need for a separate
base case, because the $n=0$ instance of the implication is the base
case, vacuously.
However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.
proof-writing induction first-order-logic
asked Jan 2 at 9:23
rb612rb612
1,848923
$endgroup$
add a comment |
$begingroup$
I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states:
Strong induction proves a sequence of statements $P(0), P(1), …$ by
proving the implication
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$,
then $P(n)$ is true."
for every nonnegative integer $n$. There is no need for a separate
base case, because the $n=0$ instance of the implication is the base
case, vacuously.
However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.
proof-writing induction first-order-logic
asked Jan 2 at 9:23
rb612rb612
1,848923
$endgroup$
2
$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32
add a comment |
$begingroup$
I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states:
Strong induction proves a sequence of statements $P(0), P(1), …$ by
proving the implication
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$,
then $P(n)$ is true."
for every nonnegative integer $n$. There is no need for a separate
base case, because the $n=0$ instance of the implication is the base
case, vacuously.
However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.
proof-writing induction first-order-logic
asked Jan 2 at 9:23
rb612rb612
1,848923
$endgroup$
I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states:
Strong induction proves a sequence of statements $P(0), P(1), …$ by
proving the implication
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$,
then $P(n)$ is true."
for every nonnegative integer $n$. There is no need for a separate
base case, because the $n=0$ instance of the implication is the base
case, vacuously.
However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.
proof-writing induction first-order-logic
proof-writing induction first-order-logic
asked Jan 2 at 9:23
rb612rb612
1,848923
asked Jan 2 at 9:23
rb612rb612
1,848923
asked Jan 2 at 9:23
rb612rb612
1,848923
asked Jan 2 at 9:23
rb612rb612
1,848923
asked Jan 2 at 9:23
rb612rb612
1,848923
1,848923
2
$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32
add a comment |
2
$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32
2
2
$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Strong (or : complete) induction is :
$(∀n)[(∀m)(m < n to P(m)) to P(n)] to (∀n) P(n)$.
So, in order to conclude with $(∀n) P(n)$ we have to show that : $(∀n)[(∀m)(m < n to P(m)) to P(n)]$ holds.
If I understand well, your concern is with $n=0$.
In that case, we have :
$(∀m)(m < 0 to P(m)) to P(0)$.
But $(m < 0 to P(m))$ is vacuously true (there are no $m < 0$). Thus, the conditional amounts to : $text T to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
add a comment |
$begingroup$
In strong induction, you show that each instance of $P(n)$ can be reduced to one or more cases $P(m)$ with $m<n$, so if all smaller cases are known to be true then case $n$ follows. The distinction from normal induction is that you don't necessarily know what values of $m$ $P(n)$ will reduce to, in particular it is not necessarily $m=n-1$, so you need to know all previous cases. A simple example is to prove that every integer at least $2$ is a product of primes: for a given integer $ngeq 2$, either $n$ is prime (so trivially true) or $n$ is a product of two integers $a,bgeq 2$. Now $a,b<n$, so both are products of primes, and we're done.
In this case there is no need for a separate base case. If $n=2$, what happens is that there are no integers $a,b$ with $2leq a,b<n$, so the second case above can't happen and $n$ must be prime. Here your reduction to smaller $m$ happens to include a proof of the base case.
However, normally what happens is that the reduction to smaller case(s) doesn't work for the smallest value of $n$, and you do need a separate base case. For example, you might be trying to prove the same statement for $ngeq 1$, and then you need to deal with $n=1$ (the empty product) separately, since the statement that $n$ is prime or can be written as the product of two smaller positive integers doesn't hold for $n=1$.
For any strong induction one of these two things happens. If the proof of the reduction breaks down for $n=0$ (or whatever the minimum $n$ is), you have to do the base case separately; if it doesn't, this gives a reason why $P(0)$ is vacuously true.
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Strong (or : complete) induction is :
$(∀n)[(∀m)(m < n to P(m)) to P(n)] to (∀n) P(n)$.
So, in order to conclude with $(∀n) P(n)$ we have to show that : $(∀n)[(∀m)(m < n to P(m)) to P(n)]$ holds.
If I understand well, your concern is with $n=0$.
In that case, we have :
$(∀m)(m < 0 to P(m)) to P(0)$.
But $(m < 0 to P(m))$ is vacuously true (there are no $m < 0$). Thus, the conditional amounts to : $text T to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
add a comment |
$begingroup$
Strong (or : complete) induction is :
$(∀n)[(∀m)(m < n to P(m)) to P(n)] to (∀n) P(n)$.
So, in order to conclude with $(∀n) P(n)$ we have to show that : $(∀n)[(∀m)(m < n to P(m)) to P(n)]$ holds.
If I understand well, your concern is with $n=0$.
In that case, we have :
$(∀m)(m < 0 to P(m)) to P(0)$.
But $(m < 0 to P(m))$ is vacuously true (there are no $m < 0$). Thus, the conditional amounts to : $text T to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
add a comment |
$begingroup$
Strong (or : complete) induction is :
$(∀n)[(∀m)(m < n to P(m)) to P(n)] to (∀n) P(n)$.
So, in order to conclude with $(∀n) P(n)$ we have to show that : $(∀n)[(∀m)(m < n to P(m)) to P(n)]$ holds.
If I understand well, your concern is with $n=0$.
In that case, we have :
$(∀m)(m < 0 to P(m)) to P(0)$.
But $(m < 0 to P(m))$ is vacuously true (there are no $m < 0$). Thus, the conditional amounts to : $text T to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
$endgroup$
Strong (or : complete) induction is :
$(∀n)[(∀m)(m < n to P(m)) to P(n)] to (∀n) P(n)$.
So, in order to conclude with $(∀n) P(n)$ we have to show that : $(∀n)[(∀m)(m < n to P(m)) to P(n)]$ holds.
If I understand well, your concern is with $n=0$.
In that case, we have :
$(∀m)(m < 0 to P(m)) to P(0)$.
But $(m < 0 to P(m))$ is vacuously true (there are no $m < 0$). Thus, the conditional amounts to : $text T to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
answered Jan 2 at 9:38
Mauro ALLEGRANZAMauro ALLEGRANZA
64.7k448112
64.7k448112
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
add a comment |
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
$begingroup$
To the proposer: So the assertion that $[forall mge 0;(m<nto P(m))]to P(n)$ holds for all $nge 0,$ does imply that the "base-case" $P(0)$ is true.
$endgroup$
– DanielWainfleet
Jan 2 at 11:36
add a comment |
$begingroup$
In strong induction, you show that each instance of $P(n)$ can be reduced to one or more cases $P(m)$ with $m<n$, so if all smaller cases are known to be true then case $n$ follows. The distinction from normal induction is that you don't necessarily know what values of $m$ $P(n)$ will reduce to, in particular it is not necessarily $m=n-1$, so you need to know all previous cases. A simple example is to prove that every integer at least $2$ is a product of primes: for a given integer $ngeq 2$, either $n$ is prime (so trivially true) or $n$ is a product of two integers $a,bgeq 2$. Now $a,b<n$, so both are products of primes, and we're done.
In this case there is no need for a separate base case. If $n=2$, what happens is that there are no integers $a,b$ with $2leq a,b<n$, so the second case above can't happen and $n$ must be prime. Here your reduction to smaller $m$ happens to include a proof of the base case.
However, normally what happens is that the reduction to smaller case(s) doesn't work for the smallest value of $n$, and you do need a separate base case. For example, you might be trying to prove the same statement for $ngeq 1$, and then you need to deal with $n=1$ (the empty product) separately, since the statement that $n$ is prime or can be written as the product of two smaller positive integers doesn't hold for $n=1$.
For any strong induction one of these two things happens. If the proof of the reduction breaks down for $n=0$ (or whatever the minimum $n$ is), you have to do the base case separately; if it doesn't, this gives a reason why $P(0)$ is vacuously true.
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
$endgroup$
add a comment |
$begingroup$
In strong induction, you show that each instance of $P(n)$ can be reduced to one or more cases $P(m)$ with $m<n$, so if all smaller cases are known to be true then case $n$ follows. The distinction from normal induction is that you don't necessarily know what values of $m$ $P(n)$ will reduce to, in particular it is not necessarily $m=n-1$, so you need to know all previous cases. A simple example is to prove that every integer at least $2$ is a product of primes: for a given integer $ngeq 2$, either $n$ is prime (so trivially true) or $n$ is a product of two integers $a,bgeq 2$. Now $a,b<n$, so both are products of primes, and we're done.
In this case there is no need for a separate base case. If $n=2$, what happens is that there are no integers $a,b$ with $2leq a,b<n$, so the second case above can't happen and $n$ must be prime. Here your reduction to smaller $m$ happens to include a proof of the base case.
However, normally what happens is that the reduction to smaller case(s) doesn't work for the smallest value of $n$, and you do need a separate base case. For example, you might be trying to prove the same statement for $ngeq 1$, and then you need to deal with $n=1$ (the empty product) separately, since the statement that $n$ is prime or can be written as the product of two smaller positive integers doesn't hold for $n=1$.
For any strong induction one of these two things happens. If the proof of the reduction breaks down for $n=0$ (or whatever the minimum $n$ is), you have to do the base case separately; if it doesn't, this gives a reason why $P(0)$ is vacuously true.
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
$endgroup$
add a comment |
$begingroup$
In strong induction, you show that each instance of $P(n)$ can be reduced to one or more cases $P(m)$ with $m<n$, so if all smaller cases are known to be true then case $n$ follows. The distinction from normal induction is that you don't necessarily know what values of $m$ $P(n)$ will reduce to, in particular it is not necessarily $m=n-1$, so you need to know all previous cases. A simple example is to prove that every integer at least $2$ is a product of primes: for a given integer $ngeq 2$, either $n$ is prime (so trivially true) or $n$ is a product of two integers $a,bgeq 2$. Now $a,b<n$, so both are products of primes, and we're done.
In this case there is no need for a separate base case. If $n=2$, what happens is that there are no integers $a,b$ with $2leq a,b<n$, so the second case above can't happen and $n$ must be prime. Here your reduction to smaller $m$ happens to include a proof of the base case.
However, normally what happens is that the reduction to smaller case(s) doesn't work for the smallest value of $n$, and you do need a separate base case. For example, you might be trying to prove the same statement for $ngeq 1$, and then you need to deal with $n=1$ (the empty product) separately, since the statement that $n$ is prime or can be written as the product of two smaller positive integers doesn't hold for $n=1$.
For any strong induction one of these two things happens. If the proof of the reduction breaks down for $n=0$ (or whatever the minimum $n$ is), you have to do the base case separately; if it doesn't, this gives a reason why $P(0)$ is vacuously true.
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
$endgroup$
In strong induction, you show that each instance of $P(n)$ can be reduced to one or more cases $P(m)$ with $m<n$, so if all smaller cases are known to be true then case $n$ follows. The distinction from normal induction is that you don't necessarily know what values of $m$ $P(n)$ will reduce to, in particular it is not necessarily $m=n-1$, so you need to know all previous cases. A simple example is to prove that every integer at least $2$ is a product of primes: for a given integer $ngeq 2$, either $n$ is prime (so trivially true) or $n$ is a product of two integers $a,bgeq 2$. Now $a,b<n$, so both are products of primes, and we're done.
In this case there is no need for a separate base case. If $n=2$, what happens is that there are no integers $a,b$ with $2leq a,b<n$, so the second case above can't happen and $n$ must be prime. Here your reduction to smaller $m$ happens to include a proof of the base case.
However, normally what happens is that the reduction to smaller case(s) doesn't work for the smallest value of $n$, and you do need a separate base case. For example, you might be trying to prove the same statement for $ngeq 1$, and then you need to deal with $n=1$ (the empty product) separately, since the statement that $n$ is prime or can be written as the product of two smaller positive integers doesn't hold for $n=1$.
For any strong induction one of these two things happens. If the proof of the reduction breaks down for $n=0$ (or whatever the minimum $n$ is), you have to do the base case separately; if it doesn't, this gives a reason why $P(0)$ is vacuously true.
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
answered Jan 2 at 9:37
Especially LimeEspecially Lime
21.8k22858
21.8k22858
add a comment |
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$begingroup$
For $n=0$, the premise ''If P(m) is true for all nonneg. integers m less than 0'' is vacuously true. But its an implication and so you need to prove that P(0) is true.
$endgroup$
– Wuestenfux
Jan 2 at 9:29
$begingroup$
@Wuestenfux Ah, so it isn't the implication that's vacuously true, it's the premise is vacuously true, so since $T Rightarrow P(0) Leftrightarrow P(0)$, it precisely boils down to proving $P(0)$?
$endgroup$
– rb612
Jan 2 at 9:32