“\(P\) only if \(Q\)” is equivalent to “if \(P\) then \(Q\)”.
Some examples of this are counterintuitive, but here’s one that feels right to me:
I go to class only if I am healthy.
Here’s the truth table:
| Row | \(P\) = “I go to class” | \(Q\) = “I am healthy” | \(P \Rightarrow Q\) |
|---|---|---|---|
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | T |
| 4 | F | F | T |
If we suppose that “I go to class only if I am healthy” is true, then only rows 1, 3, and 4 of the truth table are possible scenarios; in all of the following cases, the promise is not broken:
The only way to break the promise “I go to class only if I am healthy” is:
Biconditional - “\(P\) if and only if \(Q\)”: \(P \Leftrightarrow Q\)
This means \(P \Rightarrow Q\) and \(Q \Rightarrow P\) are both true.
In other words:
Do Exercises Parts A and B
\(\forall x\) means “for all, or”for every”, or “for each”
\(\exists\) means “there exists a” or “there is a”
Example: For every integer \(x\), \(2x\) is also an integer. Symbolically, \(\forall x \in \mathbb{Z}, 2x \in \mathbb{Z}\).
Example: There is an integer that is even. Symbolically, \(\exists x \in \mathbb{Z}\) such that \(x\) is even.
By convention, if \(P\) and \(Q\) are open sentences or statements, then \[ P \Rightarrow Q \] is a statement, with an implicit universal quantifier in front. In other words, \(P \Rightarrow Q\) means it’s impossible for \(Q\) to be false if \(P\) is true.
For example:
\(x\) is even \(\Rightarrow\) \(x\) is divisible by 6
is a true statement (we knew this already), because it’s true regardless of what \(x\) is
By this convention, the following:
\(x\) is divisble by 6 \(\Rightarrow\) \(x\) is even
is no longer interpreted as an open sentence, but it’s considered equivalent to:
\(\forall x\), \(x\) is divisble by 6 \(\Rightarrow\) \(x\) is even
which is a statement (that happens to be false).
We often have statements like “all natural numbers are positive”. We could translate this to symbols in two equivalent ways:
Do Exercises Part C