The correct option is B A→(A∨B)
ABA∧BA∨BA→BB→AA→(B→A)A→(A∧B)A→(A∨B)A→(A→B)TTTTTTTTTTTFFTFTTFTFFTFTTFTTTTFFFFTTTTTT
From the above table clearly A→(B→A)≡A→(A∨B)
Alternate Solution :
A→(B→A)⇒A→(∼B∨A)⇒∼A∨(∼B∨A)⇒∼B∨(∼A∨A)⇒∼B∨(∼A∨A)⇒∼B∨t=t (tautology)From options : A→(A∨B)⇒ ∼A∨(A∨B)⇒(∼A∨A)∨B⇒t∨B⇒t