Bourbaki

Exercise 4, § 3

Tag 03GC, permanent. Cite it as [Bourbaki, Tag 03GC].

Theory of Sets, I, 58

Let AA and BB be relations in a logical theory T\mathscr{T}. Let A∣BA|B denote the relation "(not AA) or (not BB)". Prove the following theorems in T\mathscr{T} :

(not A)⇔(A∣A),(\text{not } A) \Leftrightarrow (A|A),
(A or B)⇔((A∣A)∣(B∣B)),(A \text{ or } B) \Leftrightarrow ((A|A)|(B|B)),
(A and B)⇔((A∣B)∣(A∣B)),(A \text{ and } B) \Leftrightarrow ((A|B)|(A|B)),
(A⇒B)⇔(A∣(B∣B)).(A \Rightarrow B) \Leftrightarrow (A|(B|B)).

Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Nothing.

Transcribed from the printed volume. Not translated.