[To show that (p => q) ≡ (¬p v q) is true, we can use a truth table to demonstrate that the two expressions have the same truth values under all possible truth value assignments for p and q.

First, let's construct a truth table for (p => q) and (¬p v q):

| p | q | ¬p | ¬p v q | p => q |
|---|---|----|-------|-------|
| T | T | F | T | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | F | T |

In the truth table:
- p => q is true except when p is true and q is false.
- ¬p v q is true when either ¬p (not p) or q is true.

Observing the last two columns of the truth table, we can see that the truth values for (p => q) and (¬p v q) are the same in all cases, therefore demonstrating that (p => q) ≡ (¬p v q) is true.](URL)

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