Logique/Ensembles

f:R→R, x 7→ f(x)
f f
f f 5
f f
∃x∈R∀y∈Rx+y≥0.
z∈C
|z+ 1| ≥ 1|z−1| ≥ 1.
f:R→R
∀x∈R,∀y∈R, f(x)f(y)−f(xy) = x+y.
f:R→R
∀x∈R,∀y∈R, f(x+y) = x+f(y).
f:R→R
∀x∈R,∀y∈R, f(x−f(y)) = 1 −x−y.
E A, B, C E
A⊂B=⇒A∩C⊂B∩C A ⊂B=⇒A∪C⊂B∪C.
E A, B ∈ P(E)
A∩B=∅ ⇐⇒ A⊂Bet A∪B=E⇐⇒ B⊂A.
A∪B=A∩B A =B
A, B, C E E =A∪B∪C
D A ∩D⊂B B ∩D⊂C C ∩D⊂A
D⊂A∩B∩C
E A B E
A B
A∆B= (A∪B)\(A∩B).
A∆B= (A\B)∪(B\A)
A∆B=A∩B A =B=∅
C∈ P(E) (A∆B=A∆C)⇔(B=C)
A∆X=∅X∈ P(E)