csc165 week5 slog：
    I spent most of my study time for csc165 last weekend because of the wednesday's test. I did not do very well on my assignment and previous quizzes, so I really want to get a good test mark to regain the confidence for this course. And it worked. On monday's quiz, I felt I never feel so confident when I hand in the quiz.  I did three past mid-term tests to prepare for this test. However, because I really cared about my marks, I wrote too much explanation for the first and second questions, and I did not have enough time to consider the last question carefully and thoroughly.
The question asked me to prove that in (S3: ∃x ∈ D,(P(x) ∧ ¬Q(x)) v (¬P(x) ∧ Q(x))) and (S4:∃x∈ D,(¬P(x) v ¬Q(x))), one statement is true,and the other is false.
    My brain was in a mess, and I thought I should first do something to make the s4 looks different:
    ∃x ∈ D,(P(x) ∧ ¬Q(x)) v (¬P(x) ∧ Q(x)))
    <=> ∃x ∈ D,((P(x) ∧ ¬Q(x)) v ¬P(x)) ∧ ((P(x) ∧ ¬Q(x)) v Q(x)) #distributive law
   <=> ∃x ∈ D,((P(x)v ¬P(x)) ∧ (¬Q(x) v ¬P(x))) ∧ ((P(x) vQ(x))∧ (¬Q(x) v Q(x))) #distributive law
   <=> ∃x ∈ D, (¬Q(x) v ¬P(x)) ∧ (P(x) vQ(x)) # identity law
   <=> ∃x ∈ D, ¬(Q(x) ∧  P(x)) ∧ (P(x) v Q(x)) #
And then, I found S3: ∃x ∈ D, (¬Q(x) v ¬P(x)) ∧ (P(x) vQ(x)) and S4: ∃x∈ D,(¬P(x) v ¬Q(x))
Now, I only have to give an example that when ((P(x) vQ(x)) is false to make S3 false. And my example is : D = {1,2,3,4,5), Q = {6}, P = {4}. I am not sure whether  I am correct or not. I think I should be right because in this case, no P and Q is in D, so for all x in D,(P(x) or Q(x)) is false. And x in D is not (Q(x) and P(x)). While professor's solution is D ={1,2},P = {1}, Q = {1}. In professor's example, he proves that there is no x in D such that it is not (Q(x) and P(x)) and (P(x) or Q(x)), since the only x in P and Q is shared. The x that are not (P(x) and Q(x)) are outside P and Q, therefore, there does not exist a x in D such that (P(x) or Q(x)) is true.
     In this week's lecture, I was taught many new skills to apply on proof. For example, I can prove a statement by proving its contrapositive, since the contrapositive of a statement is actually equivalent to the statement.
    Let me set up a proof question and solve it by using contrapositive.
    The square of a even number is even, prove:
    ∀n∈N,  n^2 odd -> n odd
  
    My solution:

        ∀n∈N,  n even -> n^2 even
        Assume n∈N #n is a natural number
             Assume n is even
                Then ∃x ∈ N, n = 2x  #definition of n even
                Then n^2 = (2x)^2 = 4x^2 = 2 * 2 *x^2  # some algebra
                Then ∃y ∈ N, n^2 = 4y = 2 * (2y)  # let y = x^2 and we know x^2 must be positive
                Then ∃z ∈ N, n^2 = 2z  # let z = 2y
                Then n^2 is even  # by definition of n^2 even
            Then n is even -> n^2 is even. #from above

            Then n^2 odd -> n odd. # contrapositive
       Then ∀n∈N,  n^2 odd -> n odd  # conclusion

    Another important new idea I learnt this week is the contradiction of the contrapositive. As far as I can consider, this idea is used when the conclusion P is not very clear, so we just use the negation of assumption Q, and prove the negation of Q is wrong which is called contradiction, then we  can claim that Q is correct.