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However, you must take care of the truth in what the native says in view of whether he is a knight or a knave. Therefore, Q will involve two simple propositions: p : There is gold in island A. q : You are a knight. ” Note that Q will be a compound proposition built from p and q. In this case, if the response to your question is ‘no’, then p is false, irrespective of the value of q. Similarly, if the response is ‘yes’, then p is true, whatever the value of q may be. 5. However, the truth of Q need not be the same as the response.

If ¬ is the only connective to be used, we would generate the formulas: p0 , p1 , p2 , · · · , ¬p0 , ¬p1 , ¬p2 , . . , ¬¬p0 , ¬¬p1 , ¬¬p2 , . . , ¬¬¬p0 , ¬¬¬p1 , . . , ¬¬¬¬p0 , ¬¬¬¬p1 , . . , . . Up to equivalence, the propositions reduce to p0 , p1 , p2 , p3 , · · · , ¬p0 , ¬p1 , ¬p2 , ¬p3 , . . , . . Now, is any of these propositions equivalent to p0 ∧ p1 ? Definitely not. Because p0 ∧ p1 ≡ p0 as the interpretation i with i(p0 ) = 1, i(p1 ) = 0 is a model of p0 , but not a model of p0 ∧ p1 .

Can you see how to put commas and parentheses into this expression so that you obtain the earlier outfix notation? Taking clues from this, define a language of propositional logic without parentheses. Show that unique parsing still holds. ) 19. Construct a cnf and a dnf for the truth function u, v given by the following truth table. Simplify the normal forms and then draw circuits representing the truth functions. p 0 0 0 0 1 1 1 1 q 0 0 1 1 0 0 1 1 r 0 1 0 1 0 1 0 1 u 0 0 0 1 0 1 1 1 v 0 1 1 1 0 0 0 1 20.

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