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The simplification of[ NOT(p AND q) AND (NOT p OR q) AND (NOT q OR q) ] AND pisQuestion 4Answera.T.b.F.c.q.d.NOT q.e.None of the Given Choices.f.p.g.NOT p.

Question

The simplification of NOT(p AND q) AND (NOT p OR q) AND (NOT q OR q) AND p is:

  • a. T
  • b. F
  • c. q
  • d. NOT q
  • e. None of the Given Choices
  • f. p
  • g. NOT p.
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Solution

Sure, let's simplify the given expression step by step:

The expression is: [¬(pq)(¬pq)(¬qq)]p [ \neg(p \land q) \land (\neg p \lor q) \land (\neg q \lor q) ] \land p

  1. Simplify ¬qq\neg q \lor q: ¬qq \neg q \lor q is always true (T) because either qq is true or ¬q\neg q is true.

    So, the expression becomes: [¬(pq)(¬pq)T]p [ \neg(p \land q) \land (\neg p \lor q) \land T ] \land p

  2. Since ¬qq\neg q \lor q is true, it does not affect the AND operation: [¬(pq)(¬pq)]p [ \neg(p \land q) \land (\neg p \lor q) ] \land p

  3. Simplify ¬(pq)\neg(p \land q): ¬(pq) \neg(p \land q) is equivalent to ¬p¬q\neg p \lor \neg q (De Morgan's law).

    So, the expression becomes: [(¬p¬q)(¬pq)]p [ (\neg p \lor \neg q) \land (\neg p \lor q) ] \land p

  4. Distribute \land over \lor: (¬p¬q)(¬pq) (\neg p \lor \neg q) \land (\neg p \lor q)

    This can be simplified by considering the possible values of pp and qq:

    • If pp is true, ¬p\neg p is false, so the expression depends on qq: ¬qq \neg q \land q which is false.
    • If pp is false, ¬p\neg p is true, so the expression is true regardless of qq.

    Therefore, the expression simplifies to: ¬p \neg p

  5. Combine with the outer p\land p: ¬pp \neg p \land p

    This is always false because pp and ¬p\neg p cannot be true at the same time.

So, the final simplified expression is: False (F) \text{False (F)}

The correct answer is: b. F

This problem has been solved

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