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How do you solve (p -> q) <=> (p /\ `q -> c)?

How do you solve (p -> q) <=> (p /\ `q -> c)?

To solve this logical equivalence, we need to show that the left-hand side of the equation is logically equivalent to the right-hand side of the equation.

Here are the steps to show that (p -> q) <=> (p /\ `q -> c) is a valid logical equivalence:

  1. Start with the left-hand side of the equation: (p -> q).
  2. Use the definition of the implication operator to rewrite this as p v q, where v represents the logical OR operator.
  3. Now consider the right-hand side of the equation: (p /\ `q -> c).
  4. Use the definition of the implication operator to rewrite this as ~(p /\ ~q) v c, where ~ represents the logical NOT operator.
  5. Use De Morgan’s law to rewrite ~(p /\ ~q) as ~p v q.
  6. Substitute this into the previous equation to get (~p v q) v c.
  7. Use the associative law to rearrange the parentheses and get ~p v (q v c).
  8. Use the definition of the implication operator to rewrite this as p -> (q v c).
  9. Now we have shown that the left-hand side, (p -> q), is logically equivalent to the right-hand side, (p /\ q -> c), which we rewrote as p -> (q v c)`.

Therefore, we have shown that (p -> q) <=> (p /\ `q -> c) is a valid logical equivalence.