1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Consider Statement I: (p∧∼q)∧(∼p∧q) is a fallacy. Statement II: (p→q)↔(∼q→∼p) is a tautology.

A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

Statement I is true; Statement II is false

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Statement I is false; Statement II is true
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement IStatement I: (p∧∼q)∧(∼p∧q)≡p∧∼q∧∼p∧q≡p∧∼p∧∼q∧q≡f∧f≡f Hence, it is a fallacy statement. So, Statement I is true. Statement II (p→q)↔(∼q→∼p)≡(p→q)↔(p→q) Which is always true, so Statement II is true. Alternate Solution Statement I (p∧∼q)∧(∼p∧q) pq∼p∼qp∧∼q∼p∧q(p∧∼q)∧(∼p∧q)TTFFFF F TFFTTF F FTTFFT F FFTTFF F Hence, it is a fallacy. Statement II (p→q)↔(∼q→∼p) ∼q→∼p is contrapositive of p→q. Hence, (p→q)↔(p→q) will be a tautology

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program