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Example 2.Hey Mark,
Suppose the following are writable:
1. iiB(1)gB(2)igiB(3)gB(4)ii
2. B(5)gaB(4)
3. aiB(6)gB(1)i
Confirm the writability of iB(5)gaB(3)i.
This one is hard and to do it one must pick up on a very subtle trick. (It?s subtle in this context; if
you figure out ...[text shortened]... lain it, and feel free to discuss the questions
and share ideas.[/b]
Example 1 indicates that you want us to confirm that this symbol string:
iB(1)cB(2)iggiB(2)cB(1)i
is writable. You say that this requires showing that the following symbol string and its reverse are writable:
iB(1)cB(2)igiB(2)cB(1)i
A question: How can this symbol string be written on the last line of a derivation, given that it contradicts your introduction rule? The following is what actually ought to be derived:
iiB(1)cB(2)igiB(2)cB(1)ii
Also, in Example 2, the second premise line is as follows:
2. B(5)gaB(4)
But I don't know what to do with this line, as it isn't itself writable (it, too, violates your introduction rule). Did you intend to write the second line as follows?:
2. iB(5)gaB(4)i
Thanks,
B
Originally posted by bbarrSorry, my bad. Those lines were mistyped; the replacements you give are correct. Also, as you suggest, the following could be rules:
Hey Mark,
Example 1 indicates that you want us to confirm that this symbol string:
iB(1)cB(2)iggiB(2)cB(1)i
is writable. You say that this requires showing that the following symbol string and its reverse are writable:
iB(1)cB(2)igiB(2)cB(1)i
A question: How can this symbol string be written on the last line of a derivation, given that it co ...[text shortened]... . Did you intend to write the second line as follows?:
2. iB(5)gaB(4)i
Thanks,
B
''if iixgyiciygxii is writable, then ixggyi is writable.
if ixggyi is writable, then ixgyi and iygxi are both writable.''
Originally posted by royalchickenThis should read:
Rule 12. The Introduction Rule.
If after writing x, one can apply other rules to write y, then ixgyi is
writable.
Rule 12. The Introduction Rule
If after writing x, either from assuming it is writable or having shown it by other means, one can apply rules to write y, then ixgyi is writable.
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"Make sure you don't choose assumptions which contradict one another."
Could you elaborate on what happens if you violate this within
the NG machine? In other words, the Rules of production as
I understand them will not ever themselves yield both
x and ax.
However, if the operator of the machine chooses
B(1): The sky is blue.
B(2): a(B1) (or 'The sky is not blue', if my first def of B(2) is not acceptable)
does the machine explode?
Finally, this example shows that the final Rule and the first Rule
can find themselves in conflict if you meant to first Rule to entail
that all B() instantiations are writable by definition.
Dr. Cribs
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Originally posted by CribsProfessor Chicken tells me:
"Make sure you don't choose assumptions which contradict one another."
Could you elaborate on what happens if you violate this within
the NG machine?
Dr. Cribs
Well, if I assume x and ax, then anything I want becomes writable under the rules--my initial bible stuy thread was based on this fact. All of the B() are not a priori writable--instead, by the introduction rule, x may be assumed temporarily to show y, which shows that ixgyi is writable, and not x. So the final writability of a string cannot be assumed, but it can be assumed in order to prove a conditional.
In particular, you cannot assume x and ax, but you can assume x and derive ax (reductio ad absurdum), even though this shows that x is not ultimately writable.
~Mark
Originally posted by CribsCan you give an example of applying Rule 13?
Professor Chicken tells me:
Well, if I assume x and ax, then anything I want becomes writable under the rules--my initial bible stuy thread was based on this fact. All of the B() are not a priori writable--instead, by the introduction rule, x may be assumed temporarily to show y, which shows that ixgyi is writable, and not x. So the final writability of a s ...[text shortened]... e ax (reductio ad absurdum), even though this shows that x is not ultimately writable.
~Mark
Or is 13 really part of a Property or Theorem about
writable sequences, rather than a Rule like the rest?
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Originally posted by CribsI don't really understand--Rule 13 cannot be derived from the others. Bbarr has suggested an addendum, but I think his suggestion is derivable, so I haven't included it.
Can you give an example of applying Rule 13?
Or is 13 really part of a Property or Theorem about
writable sequences, rather than a Rule like the rest?
An example of Rule 13 in use is in:
iiiB(1)eaB(1)icaB(1)igB(1)i
which is a theorem deducible from the rules.
Originally posted by CribsRule 13 prohibits this.
Professor Chicken tells me:
if I assume x and ax, then anything I want becomes writable under the rules
Suppose I assume x and ax. Then, as you say,
y is writable, as long as y is well-formed. Also as you say,
ay is writable. But Rule 13 says that at most one of y and ay
is writable.
Unless...I am to interpret the rules to mean that starting
from x and ax, I can choose to produce either y or ay,
but once I have produced one of those, Rule 13 disallows the other
to be writable.
Dr. Cribs
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Originally posted by CribsLet's use the Introduction Rule:
Rule 13 prohibits this.
Suppose I assume x and ax. Then, as you say,
y is writable, as long as y is well-formed. Also as you say,
ay is writable. But Rule 13 says that at most one of y and ay
is writable.
Unless...I am to interp ...[text shortened]... those, Rule 13 disallows the other
to be writable.
Dr. Cribs
Assume a string and it's 'contradiction' are writable. Then, by application of other rules, the string corresponding to ''every string is writable'' is writable.
See, we've made a false assumption, given a false conclusion, and made a true conditional (since, as I wanted do do with truth tables, F->F is true). Rule 13 is what tells us that the conditional you quoted is of this type, rather than of the T->T type or the F->T type.
See, using NG as meta-NG is badass like badger.
Originally posted by royalchickenDon't you need to add to this Rule or make a new Rule 0
See, we've made a false assumption, given a false conclusion, and made a true conditional (since, as I wanted do do with truth tables, F->F is true). Rule 13 is what tells us that the conditional you quoted is of this type, rather than of the T->T type or the F->T type.
that says "Thou shalt not assume both x and ax" ?
Rule 13 says both can't be writable, but it doesn't
prohibit assuming both. But I have shown above
that if you do assume both, you can force the system
to violate Rule 13 by legally applying other Rules.
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Originally posted by CribsNo, you don't show that y is writable and ay is writable by assuming ax and x. You show that iiaxcxigiycayii is writable. This violates no rules, because it is not of the form izcazi that is prohibited.
Don't you need to add to this Rule or make a new Rule 0
that says "Thou shalt not assume both x and ax" ?
Rule 13 says both can't be writable, but it doesn't
prohibit assuming both. But I have shown above
that if you do assume both ...[text shortened]... ce the system
to violate Rule 13 by legally applying other Rules.
It's a question of proof and subproof. I really should have followed Douglas Hofstadter's example and used something like his ''fantasy rule'' which would make things more clear. More tomorrow.
''Subproof'' is not my word--it belongs to one of the long messages bbarr sent me.