DWITE '11 R3 #5 - Tautology

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Points: 10
Time limit: 1.0s
Memory limit: 64M

Problem type
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DWITE, December 2011, Problem 5

XKCD Tautology Club

We define a propositional formula as follows:

  • \{a,b,\ldots,j\} are atomic propositions, representing either true or false.
  • If A and B are propositional formulae, then so are:
    • A \wedge B\wedge is boolean "and"
    • A \vee B\vee is boolean "or"
    • \neg A\neg is boolean "not"

For example, ((a \vee b) \wedge (\neg c \vee a)) is a propositional formula. A tautology is a propositional formula that equates to true for all possible value assignments to the atomic propositions. Our previous example ((a \vee b) \wedge (\neg c \vee a)) is not a tautology because for the assignments a = \text{false}, b = \text{false} and c = \text{true}, the formula evaluates to a false. However (a \vee \neg a) is a tautology because no matter what the atomic proposition is this equates to true; (true or not-true) == true, (false or not-false) == true.

The input will contain 5 test cases, each three lines (not more than 255 characters) with a propositional formula per line.

The output will contain 5 lines of output, each three characters long. Y for tautology, N for not tautology.

Sample Input

((a v b) ^ (~c v a))
(a v ~a)
~(a ^ ~a)
a
~b
((a ^ b) v ~(c ^ ~c))

Note that ~ is used for \neg, ^ for \wedge, and v for \vee.

Sample Output

NYY
NNY

Attribution-NonCommercial-ShareAlike 3.0 Unported (CC BY-NC-SA 3.0) Problem Resource: DWITE


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