# Is it true that (P∧Q≡P)⇔(Q≡⊤)?

Consider the statement

(P∧Q≡P)⇔(Q≡⊤)

Where `P` and `Q` are statements, and `⊤` denotes the tautology (true) statement. It seems intuitively true that the above biconditional statement is true. But I would like to prove it.

One direction is easy enough to prove: suppose `Q≡⊤`; then by substitution, we have

P∧Q ≡ P∧⊤ ≡ P

by the identity law for conjunction.

However, the converse seems trickier. By assuming `P∧Q≡P`, how would one know for certain that `Q≡⊤`?

(P.S. Is it improper of me to use two different symbols for equivalence here?)

• Those symbols look a bit strange, yes. I would use a double arrow. Concerning the proof, you could simply use a truth table. Otherwise, make case distinctions over both P and Q (which is the equivalent of a truth table in natural deduction).
– user2953
Sep 29 '15 at 5:59
• Check with a truth-table : it seems not true ... Sep 29 '15 at 7:14

It is true that :

(Q≡⊤) ⊨ (P∧Q≡P)

we can check it with a truth-table.

But the other "direction" does not hold : when P and Q are both false, we have that (P∧Q≡P) is true but (Q≡⊤) is not.

As an example: P = "A person is more than 40 years old", Q = "A person is more than 30 years old". P ^ Q means "A person is both more than 40 years old and more than 30 years old", which is obviously equivalent to just P. But Q is not a tautology at all.

P∧Q≡P means that Q is true whenever P is true. Nothing more. You can't jump from this to the conclusion that Q is always true.

• Very concrete and illustrative example - helped me think about and understand it. Thanks @Gnasher729! Sep 30 '15 at 2:29

Is it true that (P∧Q≡P)⇔(Q≡⊤)?

No.

1. ((P&Q)<->P) <-> (P->Q).

2. (Q<->⊤) <-> Q.

3. [(P->Q) <-> Q] <-> (PVQ).

Therefore,

1. [(P∧Q≡P)⇔(Q≡⊤)] <-> (PvQ).