Question

∀x(P(x) ∧ Q(x)) ↔ ∀xP(x) ∧ ∀xQ(x) ◦ ∀x(P(x) ∨ Q(x)) ↔ ∀xP(x) ∨ ∀xQ(x). Please help me out with the first statement?

Answer 1

To understand these statements, we first must understand the notation being used.

• `AA` - for all - This symbol implies that something holds true for every example within a set. So, when we add a variable `x`, `AAx` means that some statement applies to every possible value or item we could substitute in for `x`.

• `P(x), Q(x)` - proposition - These are logical propositions regarding `x`, that is, they represent statements about `x` which are either true or false for any particular `x`.

• `∧` - and - This symbol allows for the combination of multiple propositions. The combined result is true when both propositions return true, and false otherwise.

• `∨` - or - This symbol also allows for the combination of multiple propositions. The combined result is false when both propositions return false, and true otherwise.

• `↔` - if and only if - This symbol also allows for the combination of multiple propositions. The combined result is true when both propositions return the same truth value for all `x`, and false otherwise.

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With this, we can now translate the statements. The first statement, directly phrased, would sound like "For all x, P of x and Q of x if and only if for all x, P of x, and for all x, Q of x."

Some minor additions and modifications makes it a little more understandable.

"For all x, P and Q are true for x if and only if P is true for all x and Q is true for all x."

This statement is a tautology, that is, it is true regardless of what we substitute in for P or Q. We can show this by demonstrating that the proposition prior to the ↔ implies the one after it, and vice versa.

Starting from the prior statement, we have that for every `x`, `P(x)∧Q(x)` is true. By our definition above, that means that for every `x`, `P(x)` is true and `Q(x)` is true. This implies that for any `x`, `P(x)` is true and for any `x`, `Q(x)` is true, which is the statement appearing after the ↔.

If we start from the statement appearing after the ↔, then we know that for any `x`, `P(x)` is true and for any `x`, `Q(x)` is true. Then for all `x`, `P(x)` and `Q(x)` are both true, meaning for all `x`, `P(x)∧Q(x)` is true. This proves that the first statement is always true.

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The second statement is false. Without going through the full process as above, we can simply show that the two propositions on either side of the ↔ do not always have the same truth value. For example, suppose that for half of all possible `x`, `P(x)` is true and `Q(x)` is false, and for the other half, `Q(x)` is true and `P(x)` is false.

In this case, as for all `x`, either `P(x)` or `Q(x)` is true, the proposition `AAx(P(x)∨Q(x))` is true (see the descriptions of ∨ above). But, because there are values for `x` for which `P(x)` is false, the proposition `AAxP(x)` is false. Similarly, `AAxQ(x)` is also false, meaning `AAxP(x)∨AAxQ(x)` is false.

As the two propositions have different truth values, clearly the truth of one does not guarantee the truth of the other, and so joining them with ↔ results in a new proposition which is false.