[rec.humor] what "cast" means

dg@wrs.UUCP (David Goodenough) (07/14/87)

In article <1987Jul9.162103.1701@sq.uucp> msb@sq.UUCP (Mark Brader) writes:
> ... lots deleted
>
>Mark Brader		"Not looking like Pascal is not a language deficiency!"
>utzoo!sq!msb							  -- Doug Gwyn

does this mean that "looking like Pascal is a language deficiency!" :-)
--
		dg@wrs.UUCP - David Goodenough

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beattie@netxcom.UUCP (Brian Beattie) (07/16/87)

In article <243@wrs.UUCP> dg@wrs.UUCP (David Goodenough) writes:
>In article <1987Jul9.162103.1701@sq.uucp> msb@sq.UUCP (Mark Brader) writes:
>> ... lots deleted
>>
>>Mark Brader		"Not looking like Pascal is not a language deficiency!"
>>utzoo!sq!msb							  -- Doug Gwyn
>
>does this mean that "looking like Pascal is a language deficiency!" :-)
I think that such a position would be defensible
>--
>		dg@wrs.UUCP - David Goodenough
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-- 
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rwhite@nu3b2.UUCP (Robert C. White Jr.) (07/17/87)

In article <243@wrs.UUCP>, dg@wrs.UUCP (David Goodenough) writes:
> In article <1987Jul9.162103.1701@sq.uucp> msb@sq.UUCP (Mark Brader) writes:
> > ... lots deleted
> >
> >Mark Brader		"Not looking like Pascal is not a language deficiency!"
> >utzoo!sq!msb							  -- Doug Gwyn
> 
> does this mean that "looking like Pascal is a language deficiency!" :-)
> --

You ever bother to learn logic? :-/


Robert.

Disclaimer:  My mind is so fragmented by random excursions into a
	wilderness of abstractions and incipient ideas that the
	practical purposes of the moment are often submerged in
	my consciousness and I don't know what I'm doing.
		[my employers certainly have no idea]

okunewck@psuvax1.psu.edu (Philip E. OKunewick) (08/06/87)

In article <243@wrs.UUCP> dg@wrs.UUCP (David Goodenough) writes:
>In article <1987Jul9.162103.1701@sq.uucp> msb@sq.UUCP (Mark Brader) writes:
>>Mark Brader		"Not looking like Pascal is not a language deficiency!"
>>utzoo!sq!msb							  -- Doug Gwyn
>
>does this mean that "looking like Pascal is a language deficiency!" :-)
>		dg@wrs.UUCP - David Goodenough

No.

We can simplify this into a mathematical proof:
(  ! = logical not;  ] = logical implication;  ^ = logical or  )

We have two basic postulates here:
1: (looking like pascal)  (l.l.p)
2: (a language deficiency)  (a l.d.)

The first statement can be written as:
!(!(looking like pascal) ] (a language deficency))
The second statement can be written as:
(looking like pascal) ] (a language deficency)
Putting it together, we get:
!((!(l.l.p.) ] (a l.d.))  ]  ((l.l.p.) ] (a l.d.))
Now, if this is the case, then we should expect:
( !(!(l.l.p.) ] (a l.d.))  ]  ((l.l.p.) ] (a l.d.)) ) = 1
replace the implication thusly: ( A ] B  =  !A ^ B )
					 (do a truth table - it works.)
(  !( !(!(l.l.p.)) ^ (a l.d.) )  ^  (!(l.l.p.) ^ (a l.d.))  ) = 1
get rid of the double negation there...
(  !( (l.l.p.) ^ (a l.d.) )  ^  (!(l.l.p.) ^ (a l.d.))  ) = 1
Now we're ready to boogie.  If the implication were correct, then we
should get a true result in 4 cases:
1: Neither looks like pascal, nor is deficient
(l.l.p. = 0, a l.d. = 0)    (  !( (0) ^ (0) )  ^  (!(0) ^ (0))  )  =  1
2: Doesn't look like pascal, but is deficient
(l.l.p. = 0, a l.d. = 1)    (  !( (0) ^ (1) )  ^  (!(0) ^ (1))  )  =  1
3: Looks like pascal but is not deficient
(l.l.p. = 1, a l.d. = 0)    (  !( (1) ^ (0) )  ^  (!(1) ^ (0))  )  =  1
4: Looks like pascal and is deficient too
(l.l.p. = 1, a l.d. = 1)    (  !( (1) ^ (1) )  ^  (!(1) ^ (1))  )  =  1
Simply each case:
1: ( !(0^0) ^ (1^0) ) = 1   (1^1) = 1  TRUE
2: ( !(0^1) ^ (1^1) ) = 1   (0^1) = 1  TRUE
3: ( !(1^0) ^ (0^0) ) = 1   (0^0) = 1  FALSE
4: ( !(1^1) ^ (0^1) ) = 1   (0^1) = 1  TRUE

So, as anybody can clearly see, one can not deduce "looking like Pascal
is a language deficiency," from the postulate "Not looking like Pascal is
not a language deficiency".

However, by the same logic, just because a language looks like Pascal
does not mean that it has major deficiencies.

Didja ever try to hack in PL/I?

						---Duck


P.S.  Fun things to tell all your friends:

A   B   !A   A ] B   !A ^ B
0   0    1     1        1
1   0    0     0        0
0   1    1     1        1
1   1    0     1        1