Saturday, April 23, 2011

How To Prove It - Chapter 1 - 1.1

1.1. Deductive Reasoning and Logical Connectives

1.
(a) A = We will have a reading assignment
B = We will have a test
C = We will have homework problems

(A ∨ C) ∧ ¬ (C ∧ B)

(b) A = you will go skiiing
B = There will be snow

¬ A ∨ (A ∧ ¬B)

(c) ¬ [(√ 7 < 2) ∨ (√ 7 = 2)]

2.
(a) A = Bill is telling the truth
B = John is telling the truth

[(A ∧ B) ∨ ¬(A ∧ B)]

(b) F = I'll have fish
C = I'll have chicken
P = I'll have mashed potatoes

(F ∧ C) ∨ (F ∧ P)

(c) S = 3 is a divisor of 6
N = 3 is a divisor of 9
F = 3 is a divisor of 15

S ∧ N ∧ F

3.
A = Alice is in the room
B = Bob is in the room
(a) (A ∧ ¬B) ∨ (¬A ∧ B)
(b) ¬(A ∧ B)
(c) ¬A ∨ ¬B
(d) ¬A ∧ ¬B

4.
(a) Yes
(b) No
(c) Yes
(d) No

5.
P = I will buy the pants
S = I will buy the shirt

(a) I will not buy the pants without the shirt
(b) I will buy neither the pants nor the shirt
(c) Either I will not buy the pants or I'll not buy the shirt

6.
S = Steve is happy
G = George is happy

(a) (S ∨ G) ∧ (¬S ∨ ¬G) - One of S and G is happy, and one of S and G is unhappy.
Simplified, One of S and G is happy and the other is unhappy.

(b) [S ∨ (G ∧ ¬S)] ∨ ¬G -
Either G is unhappy, or either G is happy but S is not or S is happy.
Another form (multisentence) -
Either G is unhappy, or one of the following is true -
a) G is happy but S is not
b) S is happy.

(c) S ∨ [G ∧ (¬S ∨ ¬G)]
Either S is happy, or G is happy and one of them is unhappy.

Note:
I'm not satisfied with my answers here.

7.
(a) JC = Jane will win the chem prize
PC = Pete will win the chem prize
JM = Jane will win the math prize
PM = Pete will win the math prize

(JM ∧ ¬PM) ∨ (¬JM ∧ PM)
PM ∨ PC
JM
Conclusion : PC

Reasoning is valid.

(b) B = beef main course
F = fish main course
P = peas
C = corn

B ∨ F
P ∨ C
¬(F ¬(B ∧ P) C)
Conclusion : ¬(B ∧ P)

Reasoning is invalid. For P = B = true, C = F = false, the arguments are true but ¬(B ∧ P) is false.

(c)
J ∨ B
¬S ∨ ¬B

Conclusion : J ∨ ¬S

Reasoning is valid.

(d) (S ∧ B) ∨ (E ∧ ¬B)

Conclusion : ¬(E ∧ S)

For S = B = E = true, the argument is valid, but the conclusion is not. Hence, reasoning is not valid.

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