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Exercise 1

1. This is a `FALSE` proposition.
2. Predicate. It is equivalent to the predicate `x > 0`.
3. This is a `FALSE` proposition.
4. This is a `TRUE` proposition.
5. This is a `FALSE` proposition.

Exercise 2

If you compare the truth tables for `A ∧ B` and `A | B`, you’ll notice a pattern:

 A B A ∧ B A | B `T` `T` `T` `F` `T` `F` `F` `T` `F` `T` `F` `T` `F` `F` `F` `T`

Whenever expression `A ∧ B` is `TRUE` , expression `A | B` is `FALSE`, and vice versa. In other words, the first expression is the negation of the second expression. This should bring you straight to the solution for expressing the `AND` in terms of the `NAND`:

`(A ∧ B) ⇔ ¬ (A | B)`

If you compare the truth ...

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