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x implies y

The title 'x implies y' is very apropos.  These are both logic problems, and a minimal understanding of implication is required.

Part I

Solution: BT - Not Frobby; WT - Frobby; BC - Frobby; WC - Not Frobby.

Since we are told the WT is frobby, we are able to deduce that the color and shape in the speaker's mind is either BT or WC.  That is, she cannot be thinking of a WT because frobbiness implies an exclusive-or of the properties, not both, and she cannot be thinking of a BC because it has neither of the properties of the frobby WT.  But each of the remaining property combinations have exactly one element from the WT.

It turns out that it doesn't matter which (of BT or WC) she actually is thinking of, however, because both WT and BC will be frobby in either case and both BT and WC will not be frobby.

Part II

Solution: two.  You must flip over the '5' card and the 'F' card.

Consider each card:

'2' - the rule says nothing about even numbers, so we needn't flip this one.

'5' - the rule requires a vowel on the opposite side of this card, so we must check it.  If there is not a vowel present, the rule is broken.

'F' - the rule's correctness also hinges on this card because if there happens to be an odd number on the back, the rule is broken.  If there is an even number, the rule remains.

'E' - we do not check this card because regardless of the number, the rule cannot be broken.  The key here is realizing the following: !((x -> y) -> (y -> x)) where '->' is read 'implies'.

Jay Miller
Thursday, September 12, 2002

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