July
01, 2017 Saturday
Bedtime
Story
Understanding the Essence of
Tonight
we will consider the idea of ω-inconsistency more formally.
Consider
arithmetic predicate ‘P’.
Then
a formal calculus C would be ω-inconsistent if it is possible to demonstrate
inside the system both these two formulas:
(1)
‘(∃x)
P(x)’ which signifies ‘There is some number that has property P’.
(2)
Each of the infinite set of formulas ‘~P(0)’, ‘~P(s0)’, ‘~P(s00)’ and so on
which signifies, ‘0 does not have the property P’, ‘1 does not have the property
P’, ‘2 does not have the property P’ and so on.
I
hope that it is almost self-evident to you that satisfying both these
conditions are contradictory and thereby making that system inconsistent or
rather ω-inconsistent.
On
one hand it is asking that some number of the system has a specific property
and on the other hand it is stating that each and every possible infinite
numbers of the system will lack that property.
A
brief analysis should make it clear that if C is inconsistent, then it is also
ω-inconsistent.
After
all, in an inconsistent formal system, all the strings within it are its
theorems.
Yet
the opposite of it may not necessarily be always true.
That
means C can be ω-inconsistent without being inconsistent.
In
which case, not only ‘(∃x)
P(x)’ may be a theorem of C but also all the other infinite set of formulas
would be theorem of C and ‘~(∃x)
P(x)’ would not be theorem of C.
Such
a scenario would allow C to be ω-inconsistent without being inconsistent (which
was proved by Gödel for his theorems).
At
first glance, it may seem contradictory for all the other infinite set of
formulas (‘~P(0)’, ‘~P(s0)’ etc.) to be theorem of C and ‘~(∃x)
P(x)’ to be not.
After
all, the whole family of formulas collectively says that there is no number
with the property P, whereas
‘~(∃x) P(x)’ singly says that no number has property P.
‘~(∃x) P(x)’ singly says that no number has property P.
Does
it not make sense for the second formula to follow directly from the first
assertion?
And
in that case, how can the formula ‘(∃x)
P(x)’ which says that some number has property P, be a theorem?
Does
it not, in that case, clearly contradict the assertion of the family of
formulas?
You
see mon ami, both these concerns are justified if you are a human ape and take
meanings into consideration.
But
C is a system of formal calculus, where meanings have no role to play.
What
holds paramount are the rules of inference.
The
concerns would be justified if there were to be one rule that could encompass
the entire family of formulas into account.
That
would have been possible if the formulas in the family were finite.
But
being infinite, no single rule can hold true for infinite number of formulas.
You
would recall how David Hilbert insisted on finitistic arithmetic in his
program.
And
so, the situation that has arisen here may seem very strange, it is still both
possible and valid.
That
is all that I will say about ω-inconsistency.
Stay tuned to the voice of an average story storytelling
chimpanzee or login at http://panarrans.blogspot.in/
Good night mon ami and my fellow cousin ape.
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Another great educator and a teacher that I am aware of is
Professor Subhashish Chattopadhyay in Bangalore, India.
While I narrate stories, Professor Subhashish an electronic
engineer and a former professor at BARC, does and teaches real mathematics and
physics.
He started the participation of Indian students at the
International Physics Olympiad.
Do visit him here:
All his books can be downloaded for free through this link:
For edutainment and English education of your children, I
recommend this large collection of Halloween Songs for Kids:
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