In the last class we explored the truth value of \[A\text{ and }B \qquad A\text{ or }B\] when we know the truth value of \(A\) and of \(B\)
Now we want to go in the other direction
Assume that we know that \[A\text{ and }B\] is true
What can we say about \(A\)? And about \(B\)?
Assume that we know that \[\text{not }(A\text{ and }B)\] is true. In other words \[A\text{ and }B\] is false.
What can we say about \(A\)? and about \(B\)?
Assume that we know that \[A\text{ or }B\] is true
What can we say about \(A\)? And about \(B\)?
We can see all possible ways to combine two statements with a double entry table like this one
F | T | |
---|---|---|
F | . | . |
T | . | . |
The first statement uses rows, the second uses columns
The logic operator depends on the four values of the table
For the “and” operator, we have
F | T | |
---|---|---|
F | F | F |
T | F | T |
Do you agree?
The table does not change if we interchange rows and columns
What does that mean?
Write the double entry table for “or”
Write the double entry table for “xor”
Consider this double entry table
F | T | |
---|---|---|
F | T | T |
T | F | T |
This table represents an operation called implication
What is the interpretation?
If is true that \(A\text{ implies }B\), what can we say about \(A\) and \(B\)
We represent this statement with the symbol \(⇒\)
We write \(A⇒B\)
F | T | |
---|---|---|
F | T | T |
T | F | T |
If \(A\) (represented in the rows) is True then \(B\) must also be True
Otherwise the statement \(A⇒B\) would be false
Notice that the matrix is not symmetric, thus the operation is not commutative
The order is important. Each part has a name
In the sentence \(A\text{ implies }B\)
If we have \[x∈P ⇒ x∈Q\] then, in set theory language, we can write \[P⊂Q\]
That is, \(P\) is a subset of \(Q\)
If \(A⇒B\) and \(B⇒A\) we can write \(A⇔B\)
We say
This is true when \(A\) and \(B\) have the same truth values.
\[A ⇒ B\] is equivalent to \[\text{not } B ⇒ \text{not } A\]
\[A ⇒ B\] is equivalent to \[(\text{not } A )\text{ or }B\]
\[\text{not }(A ⇒ B)\] is equivalent to \[A \text{ and }\text{not } B\]
\[A ⇒ (B ⇒ C)\] is equivalent to \[(A \text{ and } B) ⇒ C\]
\[\left(A ⇒ (B ⇒ C)\right)\] is equivalent to \[\left(B ⇒ (A ⇒ C)\right)\]
\[\left(C ⇒ (A ⇒ B)\right)\] is equivalent to \[\left((C ⇒ A) ⇒ (C ⇒ B)\right)\]
In math we look for theorems. They all are like
If A is true and B is true then C is true
In other words, theorems are implications
To prove that a theorem is true, we need to see all the small implications
The axioms are statements that we assume to be true. They are the starting point of theorems
Euclid believed these assumptions were intuitive
From these axioms, Euclid proved that
This is valid in particular for triangles and squares, because
If is true that
Then (8) and (9) and (10) imply that \[2ab+c^2 = a^2+2ab+b^2\] and therefore \[c^2 = a^2+b^2\]
If \(A⇒B\) and we can show that \(B\) is false, we have proved that \(A\) is false. There is a famous example
Pythagoras theorem says that a right triangle with sides of length 1 has an hypotenuse of length \(\sqrt{2}\)
Greeks believed that all numbers were rational (i.e. like \(p/q\))
A student of Pythagoras used a proof by contradiction to show that \(\sqrt{2}\) cannot be rational (i.e. cannot be written as a fraction)
The legend says that his classmates killed him because of this
Assume that \(\sqrt{2}=p/q\) with \(p\) and \(q\) natural numbers
Assume also that \(p\) and \(q\) don’t have common factors
Squaring both sides, we have \(2q^2=p^2\)
Therefore \(p^2\) is even. Thus \(p\) is even, and \(p=2r\)
Thus, \(2q^2=4 r^2\), which implies \(q^2=2r^2\)
Therefore \(q\) is even. And (4) says that \(p\) is also even
But this cannot be true, because of (2)
Therefore (1) must be false. \(\sqrt{2}\) cannot be a fraction
An argument is a phrase saying that IF several predicates (called premises) are true, THEN another predicate (called conclusion) must also be true
All the premises are connected by AND
The argument is valid if it is a tautology
If the argument is not correct, we say it is a fallacy
These examples are taken from The Internet Encyclopedia of Philosophy
\[((Honda \text{ or } Toyota) \text{ and } \text{not } Honda) ⇒ Toyota\]
Notice that the argument is valid even if Elizabeth as no car
\[\begin{aligned}(Toaster ⇒ Gold) \text{ and } (Gold⇒ TimeMachine)\\ ⇒ (Toaster ⇒ TimeMachine)\end{aligned}\]
The first two propositions are not true
Nevertheless if they were true, the third proposition is necessarily true
An argument may be valid even if the premises are never true
A argument is sound if and only if it is both valid, and all of its premises are actually true.
An invalid argument is a logical fallacy. Be aware of them!
See also:
We saw that \((P\vee Q) \text{ and } \text{not } P) ⇒ Q\) and \[(P⇒ Q) \text{ and } (Q⇒ R)⇒ (P⇒ R)\] We also have “modus ponens” \[((P⇒ Q) \text{ and } P)⇒ Q\] and “modus tollens” \[((P⇒ Q) \text{ and } \text{not } Q)⇒ \text{not } P\]
If being rich makes you happy and you are rich, then you are happy
IF you being rich IMPLIES you being happy AND you are rich THEN you are happy
IF \(Rich⇒ Happy\) AND \(Rich\), THEN \(Happy\)
\[((P⇒ Q) \text{ and } P)⇒ Q\]
If being rich makes you happy and you are unhappy, then you are not rich
IF you being rich IMPLIES you being happy AND you are not happy THEN you are not rich
IF \(Rich⇒ Happy\) AND \(\text{not } Happy\), THEN \(\text{not } Rich\)
\[((P⇒ Q) \text{ and } \text{not } Q)⇒ \text{not } P\]