Metamath Proof Explorer


Theorem dnnumch2

Description: Define an enumeration (weak dominance version) of a set from a choice function. (Contributed by Stefan O'Rear, 18-Jan-2015)

Ref Expression
Hypotheses dnnumch.f ⊢ F = recs ⁡ z ∈ V ⟼ G ⁡ A ∖ ran ⁡ z
dnnumch.a ⊢ φ → A ∈ V
dnnumch.g ⊢ φ → ∀ y ∈ 𝒫 A y ≠ ∅ → G ⁡ y ∈ y
Assertion dnnumch2 ⊢ φ → A ⊆ ran ⁡ F

Proof

Step Hyp Ref Expression
1 dnnumch.f ⊢ F = recs ⁡ z ∈ V ⟼ G ⁡ A ∖ ran ⁡ z
2 dnnumch.a ⊢ φ → A ∈ V
3 dnnumch.g ⊢ φ → ∀ y ∈ 𝒫 A y ≠ ∅ → G ⁡ y ∈ y
4 1 2 3 dnnumch1 ⊢ φ → ∃ x ∈ On F ↾ x : x ⟶ 1-1 onto A
5 f1ofo ⊢ F ↾ x : x ⟶ 1-1 onto A → F ↾ x : x ⟶ onto A
6 forn ⊢ F ↾ x : x ⟶ onto A → ran ⁡ F ↾ x = A
7 5 6 syl ⊢ F ↾ x : x ⟶ 1-1 onto A → ran ⁡ F ↾ x = A
8 resss ⊢ F ↾ x ⊆ F
9 rnss ⊢ F ↾ x ⊆ F → ran ⁡ F ↾ x ⊆ ran ⁡ F
10 8 9 mp1i ⊢ F ↾ x : x ⟶ 1-1 onto A → ran ⁡ F ↾ x ⊆ ran ⁡ F
11 7 10 eqsstrrd ⊢ F ↾ x : x ⟶ 1-1 onto A → A ⊆ ran ⁡ F
12 11 a1i ⊢ φ → F ↾ x : x ⟶ 1-1 onto A → A ⊆ ran ⁡ F
13 12 rexlimdvw ⊢ φ → ∃ x ∈ On F ↾ x : x ⟶ 1-1 onto A → A ⊆ ran ⁡ F
14 4 13 mpd ⊢ φ → A ⊆ ran ⁡ F