Tuesday, April 13, 2010

how to prove it - ch1, sec1.1 ex

P $\equiv$ We have a reading assignment.
Q $\equiv$ We have homework.
R $\equiv$ We have a test.

$(P \lor Q) \land \lnot(Q \land R)$

P $\equiv$ You will go skiing.
Q $\equiv$ There will be snow.

$\lnot P \lor (P \land \lnot Q)$

P $\equiv \sqrt{7} < 2$
Q $\equiv \sqrt{7} = 2$

$\lnot (P \lor Q)$

J $\equiv$ John is telling the truth.
B $\equiv$ Bill is telling the truth.

$(J \land B) \lor (\lnot J \land \lnot B)$

F $\equiv$ I'll have fish.
C $\equiv$ I'll have chicken.
P $\equiv$ I'll have mashed potatoes.

$(F \lor C) \land \lnot (F \land P)$

P $\equiv$ 3 is a common divisor of 6.
Q $\equiv$ 3 is a common divisor of 9.
R $\equiv$ 3 is a common divisor of 15.

$P \land Q \land R$

Ex-3(a) $(A \lor B) \land \lnot (A \land B)$
Ex-3(b) $\lnot A \land \lnot B$
Ex-3(c) $\lnot A \lor \lnot B$
Ex-3(d) $\lnot A \land \lnot B$

Ex-4 a and c only

$\lnot (P \land \lnot S) = \lnot P \lor S$ means: Either I will not buy the pants or I'll buy the shirt
Ex-5(b) I will neither buy pants nor shirt.
Ex-5(c) Either I'll not buy pants or I'll not buy the shirt... Or we could say... I'll not buy both pants and shirt.

Jane and Pete won't both win the math prize. $\equiv \lnot (Jm \land Pm)$
Pete will either win the math prize or the chemistry prize. $\equiv (Pm \lor Pc)$
Jane will win the math prize. $\equiv Jm$

Pete will win the chemistry prize. $\equiv Pc$

Reasoning is valid.

The main course will be either beef or fish. $\equiv B \lor F$
The vegetable will be either peas or corn. $\equiv P \lor C$
We will not have both fish as a main course and corn as a vegetable. $\equiv \lnot (F \land C)$

We will not have both beef as a main course and peas as a vegetable. $\equiv \lnot (B \land P)$

Reasoning is not valid because when B = true, P = true, F = false, C = false; all the premises are true but the conclusion is false.

Either John or Bill is telling the truth. $\equiv J \lor B$
Either Sam or Bill is lying. $\equiv \lnot S \lor \lnot B$

Either John is telling the truth or Sam is lying. $\equiv J \lor \lnot S$

Reasoning is valid.

Either sales will go up and the boss will be happy, or expenses will go up and the boss won't be happy. $\equiv (S \land B) \lor (E \land \lnot B)$

Sales and expenses will not both go up. $\equiv \lnot (S \land E)$

Reasoning is not valid because when S = true, B = true, E = true; the premise is true but conclusion is false.


  1. For 3a) I have ¬(A∧B):
    (A∧B) - both are in the room
    ¬(A∧B) - not both in the room

    (A∨B)∧¬(A∧B) - I think you've constructed an "exclusive or" so one of them must be in the room. I think the statement "Alice and Bob are not both in the room" admits the possibility of neither being in the room.

  2. @charlieb yes, I've tried to ensure that exactly one person is in the room but not both. English can be ambiguous, so probably both of us are correct in our own interpretations.

  3. Yeah for 3(a) I got (¬A ∨ ¬B) which is same as cheliesb's answer. But as you said, it could be argued both ways I guess.

  4. I am having trouble with question 7(d) of section 1.1
    What I don't understand is why you said that when S is true, E is true. Does an increase in sales NECESSARILY imply an increase in expenses? Or do you mean that since S is true, E CAN be true because there is no restriction that says "not"(E and B)

  5. I agreed with you until I did a truth table and now agree with Himansh that conclusion is false.