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Use a truth table to show that p⇒q is logically equivalent to ¬p∨q.
I’ll get you started: Fill out the following truth table:
| p | q | ¬p | ¬p∨q | p⇒q | ¬p∨q⇔p⇒q |
|---|---|---|---|---|---|
| T | T | ? | ? | ? | ? |
| T | F | ? | ? | ? | ? |
| F | T | ? | ? | ? | ? |
| F | F | ? | ? | ? | ? |
Note that you can type that sort of thing into the forum using standard Markdown table syntax. Here’s the Markdown that I typed in to generate the table above. You’ve just got to figure out what to replace the question marks with.
|p|q|¬p|¬p∨q|p⇒q|¬p∨q⇔p⇒q|
|:---:|:---:|:---:|:---:|:---:|:---:|
|T|T| ? | ? | ? | ? |
|T|F| ? | ? | ? | ? |
|F|T| ? | ? | ? | ? |
|F|F| ? | ? | ? | ? |
