We will deal with statements that are either True or False
For example
Sometimes they are also called propositions
In math, when we want to make things general, we get rid of the “noise”, that is, the details that make things complicated and are not relevant to us.
In this case we are going to represent statements with uppercase letters
We can change and modify statements to make new ones
We can say
The simplest way to change a statement is to negate it
For instance, if \[A\] means “It is raining”, then \[\text{not }A\] means “it is not raining”
There are several ways of combining two statements
The two more common are \[A\text{ and }B \qquad A\text{ or }B\]
The first one is sometimes called conjunction
The second one is called disjunction
As you can guess, \[A\text{ and }B\] is true when both \(A\) and \(B\) are True
This is an important tool
We can write all combinations of \(A\) and \(B\) and evaluate the value of \(A\text{ and }B\) in each case
\(A\) | \(B\) | \(A\text{ and }B\) |
---|---|---|
F | F | F |
F | T | F |
T | F | F |
T | T | T |
We will make many truth tables
Exercise: How many rows are needed for any truth table?
It is easy to see that \[A\text{ and }B = B\text{ and }A\] (i.e. conjunction is commutative) \[(A\text{ and }B)\text{ and }C = A\text{ and }(B\text{ and }C)\] (i.e. conjunction is associative)
Since conjunction is associative we can write \[A\text{ and }B\text{ and }C\] without parenthesis, since there is no ambiguity
This expression will be true when all the statements are True
Exercise: write the truth table so you can convince that conjunction is commutative and associative
As you can guess, \[A\text{ or }B\] is true when either \(A\) or \(B\) are True
Exercise: Write the truth table for disjunction
Everyday language can be ambiguous. We may say
In the first case you can do both.
In the second one you can only choose one
The last one is called exclusive or (xor)
Exercise: Write the truth table for xor
It is easy to see that \[A\text{ or }B = B\text{ or }A\] (i.e. disjunction is commutative) \[(A\text{ or }B)\text{ or }C = A\text{ or }(B\text{ or }C)\] (i.e. disjunction is associative)
Since disjunction is associative we can write \[A\text{ or }B\text{ or }C\] without parenthesis, since there is no ambiguity
This complex expression will be true when any of the statements is True
Notice that \[(A\text{ and }B)\text{ or }C\] is not the same as \[A\text{ and }(B\text{ or }C)\]
Thus, we need parenthesis when we combine “and” with “or”
Exercise: Write the truth table and verify that they are different
So, what is \(A\text{ and }(B\text{ or }C)\)?
This is a little like \(A⋅(B+C)\)
which is equal to \(A⋅B + A⋅C\)
so we would expect a similar formula
Indeed, if we build the truth table, we can see that \[A\text{ and }(B\text{ or }C) = (A\text{ and }B)\text{ or }(A\text{ and }C)\]
Exercise: Write the truth table and verify that they are the same
We have seen that \[A\text{ and }(B\text{ or }C) = (A\text{ and }B)\text{ or }(A\text{ and }C)\] which is similar to the formulas with numbers
But these are not numbers, so there are different formulas
For instance we also have \[A\text{ or }(B\text{ and }C) = (A\text{ or }B)\text{ and }(A\text{ or }C)\]
We can write \[\text{not }(A\text{ and }B)\] and see that it is different from \[(\text{not }A)\text{ and }B\] Again, we must be careful and use parenthesis
Exercise: Write the truth table and verify that they are different
These two rules are super powerful
\[\begin{aligned} \text{not }(A\text{ and }B) &= (\text{not }A)\text{ or }(\text{not }B)\\ \text{not }(A\text{ or }B) &= (\text{not }A)\text{ and }(\text{not }B) \end{aligned}\]
In English
I assume that you know the basic ideas of sets
For any set \(P\), the expression \[x∈P\] is a logical statement. Either true or false. Then \[\begin{aligned} x\not∈ P &= \text{not } (x∈P)\\ x∈P∩Q &= (x∈P)\text{ and } (x∈Q)\\ x∈P∪Q &= (x∈P)\text{ or } (x∈Q)\\ \end{aligned}\]