Week 3: Conjunction, Disjunction, Negation

In week 3, Danny taught the concept of conjunction, disjunction, and implication. I learnt these new learning by reviewing the material consequently. What I really found easy and got surprised during week 3 was how the course material tied up with Venn diagram question using 'or' and 'and'. Using the word, 'or', I can make a conjunction (also known as intersection). Whereas using the word 'and', I can simply find the disjunction.

The following symbols are used to represent conjunction and disjunction.  

Type	Meaning	Symbol
Disjunction	Or	∨
Conjunction	And	∧

The next important concept I learnt was the concept of logical negation. Danny taught the class that not not can be reinterpreted as negation of negation meaning there is no change from the original example. For instance, below example below deals with concept of negation.

Sam: "I am not not UofT student".

Here, the above sentence describes 'I am UofT student' as two 'not' cancels with each other. However, in our real life, we don't use word 'not not', and the reason behind this may be to make a sentence simpler. This also goes true when I am constructing the code in computer programming.

After that, in the standard negation slide in the PowerPoint I learnt the following negation means the same to each other.

¬∀x∈ X,... ↔ ∃x∈ X, ¬...	¬(P(x) → Q(x)) ↔ P(x) ∧ ¬Q(x)
¬∃x∈ X,... ↔ ∀x∈ X, ¬...	P(x) ∧ ¬Q(x) ↔ ¬(P(x) → Q(x))

The following below standard negation was what I found the most hard. 

 ¬(P(x) → Q(x)) ↔ P(x) ∧ ¬Q(x)

This can be interpreted as not (if p is true, then Q is also true) is the same with intersection of P(x) and  not Q(x). Later, Danny showed us the truth table which I learnt in week 2: implication. The concept of implication being false is the only case when if P is true, then Q is false. Using this knowledge as our basis to truth table,  I was able to answer how the relation between P,Q, and R (which is the third variable defined by Danny) depending on the situation. 

Mathematical approach I came up with for this week is to change the below equation without using conjunction and negation.  

¬((¬P(x) → Q(x)) ∧ R(x)) → S(x))

This can be interpreted as Not (Not(if P(x) is true, then Q(x) is also true) and R(x)) is true, then S(x) is true)

First I took out the negation which is placed at the beginning using 
¬(P(x) → Q(x)) ↔ P(x) ∧ ¬Q(x).

Then new equation would be (¬P(x) → Q(x)) ∧ R(x)) → ¬S(x)

Then, (¬P(x) → Q(x)) ∧ R(x)) can be simplified into ¬(R(x) → (P(x) → Q(x))).

Now, ¬(R(x) → (P(x) → Q(x))) → ¬S(x) can be come up without using conjunction. This can be interpreted as "not[(if R(x) is is true, then (if P(x) is true, then Q(x) is true) is also true )] is true, then not(S(x) is also true." 

And as two "not" is used, even if I cancel "not" and restate the following statement as 
(R(x) → (P(x) → Q(x))) → S(x), the statement still contains the same information as 
¬((¬P(x) → Q(x)) ∧ R(x)) → S(x)). 

Week 3 seems to be a bit more difficult than week 2, but the course material is fairly consistent with the Slog writing enabling me to remind myself what I learnt during the week.