Metamath Proof Explorer


Theorem fseqdom

Description: One half of fseqen . (Contributed by Mario Carneiro, 18-Nov-2014)

Ref Expression
Assertion fseqdom ⊢ A ∈ V → ω × A ≼ ⋃ n ∈ ω A n

Proof

Step Hyp Ref Expression
1 omex ⊢ ω ∈ V
2 ovex ⊢ A n ∈ V
3 1 2 iunex ⊢ ⋃ n ∈ ω A n ∈ V
4 xp1st ⊢ x ∈ ω × A → 1 st ⁡ x ∈ ω
5 peano2 ⊢ 1 st ⁡ x ∈ ω → suc ⁡ 1 st ⁡ x ∈ ω
6 4 5 syl ⊢ x ∈ ω × A → suc ⁡ 1 st ⁡ x ∈ ω
7 xp2nd ⊢ x ∈ ω × A → 2 nd ⁡ x ∈ A
8 fconst6g ⊢ 2 nd ⁡ x ∈ A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x : suc ⁡ 1 st ⁡ x ⟶ A
9 7 8 syl ⊢ x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x : suc ⁡ 1 st ⁡ x ⟶ A
10 9 adantl ⊢ A ∈ V ∧ x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x : suc ⁡ 1 st ⁡ x ⟶ A
11 elmapg ⊢ A ∈ V ∧ suc ⁡ 1 st ⁡ x ∈ ω → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ A suc ⁡ 1 st ⁡ x ↔ suc ⁡ 1 st ⁡ x × 2 nd ⁡ x : suc ⁡ 1 st ⁡ x ⟶ A
12 6 11 sylan2 ⊢ A ∈ V ∧ x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ A suc ⁡ 1 st ⁡ x ↔ suc ⁡ 1 st ⁡ x × 2 nd ⁡ x : suc ⁡ 1 st ⁡ x ⟶ A
13 10 12 mpbird ⊢ A ∈ V ∧ x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ A suc ⁡ 1 st ⁡ x
14 oveq2 ⊢ n = suc ⁡ 1 st ⁡ x → A n = A suc ⁡ 1 st ⁡ x
15 14 eliuni ⊢ suc ⁡ 1 st ⁡ x ∈ ω ∧ suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ A suc ⁡ 1 st ⁡ x → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ ⋃ n ∈ ω A n
16 6 13 15 syl2an2 ⊢ A ∈ V ∧ x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ ⋃ n ∈ ω A n
17 16 ex ⊢ A ∈ V → x ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x ∈ ⋃ n ∈ ω A n
18 nsuceq0 ⊢ suc ⁡ 1 st ⁡ x ≠ ∅
19 fvex ⊢ 2 nd ⁡ x ∈ V
20 19 snnz ⊢ 2 nd ⁡ x ≠ ∅
21 xp11 ⊢ suc ⁡ 1 st ⁡ x ≠ ∅ ∧ 2 nd ⁡ x ≠ ∅ → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x = suc ⁡ 1 st ⁡ y × 2 nd ⁡ y ↔ suc ⁡ 1 st ⁡ x = suc ⁡ 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y
22 18 20 21 mp2an ⊢ suc ⁡ 1 st ⁡ x × 2 nd ⁡ x = suc ⁡ 1 st ⁡ y × 2 nd ⁡ y ↔ suc ⁡ 1 st ⁡ x = suc ⁡ 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y
23 xp1st ⊢ y ∈ ω × A → 1 st ⁡ y ∈ ω
24 peano4 ⊢ 1 st ⁡ x ∈ ω ∧ 1 st ⁡ y ∈ ω → suc ⁡ 1 st ⁡ x = suc ⁡ 1 st ⁡ y ↔ 1 st ⁡ x = 1 st ⁡ y
25 4 23 24 syl2an ⊢ x ∈ ω × A ∧ y ∈ ω × A → suc ⁡ 1 st ⁡ x = suc ⁡ 1 st ⁡ y ↔ 1 st ⁡ x = 1 st ⁡ y
26 sneqbg ⊢ 2 nd ⁡ x ∈ V → 2 nd ⁡ x = 2 nd ⁡ y ↔ 2 nd ⁡ x = 2 nd ⁡ y
27 19 26 mp1i ⊢ x ∈ ω × A ∧ y ∈ ω × A → 2 nd ⁡ x = 2 nd ⁡ y ↔ 2 nd ⁡ x = 2 nd ⁡ y
28 25 27 anbi12d ⊢ x ∈ ω × A ∧ y ∈ ω × A → suc ⁡ 1 st ⁡ x = suc ⁡ 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y ↔ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y
29 22 28 bitrid ⊢ x ∈ ω × A ∧ y ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x = suc ⁡ 1 st ⁡ y × 2 nd ⁡ y ↔ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y
30 xpopth ⊢ x ∈ ω × A ∧ y ∈ ω × A → 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x = 2 nd ⁡ y ↔ x = y
31 29 30 bitrd ⊢ x ∈ ω × A ∧ y ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x = suc ⁡ 1 st ⁡ y × 2 nd ⁡ y ↔ x = y
32 31 a1i ⊢ A ∈ V → x ∈ ω × A ∧ y ∈ ω × A → suc ⁡ 1 st ⁡ x × 2 nd ⁡ x = suc ⁡ 1 st ⁡ y × 2 nd ⁡ y ↔ x = y
33 17 32 dom2d ⊢ A ∈ V → ⋃ n ∈ ω A n ∈ V → ω × A ≼ ⋃ n ∈ ω A n
34 3 33 mpi ⊢ A ∈ V → ω × A ≼ ⋃ n ∈ ω A n