7

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≡⊤?

Thanks in advance for your thoughts.

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

  • 2
    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). – Keelan Sep 29 '15 at 5:59
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    Check with a truth-table : it seems not true ... – Mauro ALLEGRANZA Sep 29 '15 at 7:14
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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.

3

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! – Mathemanic Sep 30 '15 at 2:29
0

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).

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