GO Classes CS 2025 | Weekly Quiz 3 | Propositional Logic | Question: 8

Answer 1

Detailed Video Solution: https://youtu.be/KFVdMzQX_sc 

Now, we can symbolically express the two assumptions:

1. $(D(l) \vee \neg S(l))$
2. $(E(m) \rightarrow \neg D(l))$

Observe that (by the definition of implication and by contraposition, respectively) these two are equivalent to the following:

1. $(S(l) \rightarrow D(l))$
2. $(D(l) \rightarrow \neg E(m))$

We are now ready to check the following claims:
 

• A. "Mathematics is not easy if many students like logic." This claim can be written as $(S(l) \rightarrow \neg E(m))$. Since we know from above that $(S(l) \rightarrow D(l))$, and that $(D(l) \rightarrow \neg E(m))$, we conclude that this claim is true.
   
• B. "Not many students like logic if mathematics is not easy." This claim can be written as $(\neg E(m) \rightarrow \neg S(l))$. By contraposition, we obtain $(S(l) \rightarrow E(m))$. Again, we know from our hypotheses that $(S(l) \rightarrow D(l))$, and that $(D(l) \rightarrow \neg E(m))$, so we conclude that this claim is false.
   
• C. "Mathematics is not easy or logic is difficult." We can write this claim as $(\neg E(m) \vee D(l))$, which is equivalent (by the definition of implication) to $(E(m) \rightarrow D(l))$. However, this claim contradicts $(E(m) \rightarrow \neg D(l))$, which is one of our hypotheses. Therefore the claim is false.
   
• D. "Logic is not difficult or mathematics is not easy." Symbolically, we express this claim as $(\neg D(l) \vee \neg E(m))$. If we convert this to an implication, we get $(D(l) \rightarrow \neg E(m))$, which is one of our hypotheses. The claim is therefore true.

References:

• http://www.cs.cornell.edu/courses/cs280/2002sp/homework/hw02sol.pdf
• https://gateoverflow.in/306902/kenneth-rosen-edition-7-exercise-1-6-question-34-page-no-80

Answer 2

Statement 1 of this question can be easily misinterpreted during exam pressure (talking with personal experience here :P)

Statement 1: Logic is difficult or not many students like logic

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Misinterpretation: (Logic is difficult or not, still many students like logic)

This interpretation will lead to the wrong answer. With this interpretation Option B will be the correct answer which is not. So. Be cautious with this type of questions.

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Back to the Correct Answer:

$L: Logic\ is\ difficult.$

$S: Many\ Students\ like\ logic.$

$M: Maths\ is\ easy.$

Now, this is the correct interpretation of the statements:

$P1: L\ \cup \sim S$

$P2: M\rightarrow\sim L$

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Option A: True

S -> ~M

Let this be False and try to check if P1 and P2 can be made True. (Analysis of Implication)

Therefore, M = T, S = T
Substituting this in P1 and P2

P1: L + F
P2: T -> ~L

Case 1: L = F
P1 = False

Case 2: L = T
P1 = T
P2 = F

In both the cases, Premises cannot be made True,
Therefore, Option A is a valid conclusion.
 

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Option B: False

~M -> ~S

For this to be False,
S = T, M = F

Substituting in P1 and P2:

P1: L + F
P2: F -> ~L

P2 = True
P1 = True, when L = True

Therefore, Option B is not a valid conclusion.

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Option C: False

~M + L

For this to be False,
M = T, L = F

Substituting this in P1 and P2:

P1: F + ~S
P2: T -> T

P2 = True
P1 = True, when S = False

Therefore, Option C is not a valid conclusion.

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Option D: True

~L + ~M

For this to be False,
L = T, M = T

Substituting this in P1 and P2:

P1: T + ~S
P2: T -> F

P2 = False, always
P1 does not matter here, because (P1).(P2) = False

Therefore, Option 4 is a valid conclusion.

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Note:

Analysis of Impact says that 
P -> Q will be False
only when P is True and Q is False.

For all the other cases, P -> Q = True

So, if

P is a set of Premises ($P: P1\wedge P2\wedge P3...\wedge Pn$)

and Q is the Conclusion

So, if we can prove this logical Argument, that is (P -> Q) False somehow, then the conclusion is invalid
But if we cannot, then the conclusion is valid