Explanation
Statement (S1) ka Analysis
(S1):(p⇒q)∧(p∧(∼q))
-
Hame pata hai ki (p⇒q) logically equivalent hota hai (∼p∨q) ke.
-
Ab expression ko aise likh sakte hain: (∼p∨q)∧(p∧∼q).
-
Dhyan se dekhein toh (p∧∼q) darasal ∼(∼p∨q) hai (De Morgan's Law).
-
Toh equation bani: X∧(∼X), jahan X=(p⇒q).
-
Kisi bhi statement ka uske negation ke saath "AND" (∧) hamesha Contradiction (False) hota hai.
Statement (S2) ka Analysis
(S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q))
Isse simplify karne ke liye distributive law ka use karte hain:
-
Pehle do terms: ((p∨∼p)∧q)=(T∧q)=q.
-
Aakhri do terms: ((p∨∼p)∧∼q)=(T∧∼q)=∼q.
-
Ab bacha: q∨∼q.
-
q∨∼q hamesha Tautology (True) hota hai.
Final Answer
Chunkani dono statements sahi hain, isliye sahi option (3) both are true hai.
