Fc. ∀t,E, Fnc (E ∋ x ↦ t(x))
1. ∀t,E, ∀Fnc f, f = (E
∋ x ↦ t(x)) ⇔ (Dom f = E ∧
(∀x∈E, f(x) = t(x)))
1'. ∀t,E, ∀f, f = (E ∋ x ↦ t(x))
⇔ (Fnc f ∧ Dom f = E ∧ ∀x∈E, f(x) = t(x))
2. ∀t,E, Dom(E ∋ x ↦ t(x)) = E
∧ ∀y∈E, (E ∋ x ↦ t(x))(y) = t(y)
(=Fnc) ∀Fnc f, ∀Fnc g,
(Dom f = Dom g ∧ ∀x∈Dom f, f(x)
= g(x)) ⇒ f = g
1a. ∀t,E, ∀Fnc f, f = (E
∋ x ↦ t(x)) ⇒ (Dom f = E ∧
(∀x∈E, f(x) = t(x)))
1b. ∀t,E, ∀Fnc f, (Dom f = E ∧
(∀x∈E, f(x) = t(x))) ⇒
f = (E ∋ x ↦ t(x))
We can see immediately that 1a'. ⇔ (Fc. ∧ 2.) ⇒ 1a.
(We may write 2. ⇒ 1a. if not careful about definiteness).
Conversely, we can see (1a. ∧ Fc.) ⇒ 2.
∀Fnc f, ∀Fnc g, (Dom f = Dom g ∧ ∀x∈Dom f, f(x) = g(x)) ⇒ ∃SetE, Dom f = E = Dom g ∧ (∀x∈E, f(x) = f(x) = g(x)) ∴ f = (E ∋ x ↦ f(x)) = g.Proof of (2.∧(=Fnc)) ⇒ 1b. : ∀t,E, ∀Fnc f,
(Dom f = E ∧ (∀x∈E, f(x) = t(x)))
⇒ (Dom f = Dom(E ∋ x ↦ t(x)) ∧ (∀y∈E, f(y) = (E ∋ x ↦ t(x))(y)))
⇒ f = (E ∋ x ↦ t(x))