Let p and q be two statements. Consider the following logical identities: (A)…

Check each statement:

(A) ∼(p ∧ q) = ∼p ∨ ∼q
This is De Morgan's law, so it is true.

(B) ∼(p ∨ q) = ∼p ∧ ∼q
This is also De Morgan's law, so it is true.

(C) p ∧ ∼p = T
A statement and its negation cannot both be true. Therefore p ∧ ∼p = F, not T. So (C) is false.

(D) ∼(p → q) = p ∧ ∼q
Since p → q ≡ ∼p ∨ q, we get:
∼(p → q) = ∼(∼p ∨ q) = p ∧ ∼q
So (D) is true.

(E) p ∨ q = ∼p ∨ ∼q
This is not a valid logical identity. For example, if p = T and q = T, then p ∨ q = T but ∼p ∨ ∼q = F. So (E) is false.

Therefore, the correct statements are (A), (B), and (D).

Final Answer: (A), (B) and (D) Only.