Metamath Proof Explorer


Theorem 2lgslem1b

Description: Lemma 2 for 2lgslem1 . (Contributed by AV, 18-Jun-2021)

Ref Expression
Hypotheses 2lgslem1b.i ⊢ I = A … B
2lgslem1b.f ⊢ F = j ∈ I ⟼ j ⋅ 2
Assertion 2lgslem1b ⊢ F : I ⟶ 1-1 onto x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2

Proof

Step Hyp Ref Expression
1 2lgslem1b.i ⊢ I = A … B
2 2lgslem1b.f ⊢ F = j ∈ I ⟼ j ⋅ 2
3 eqeq1 ⊢ x = j ⋅ 2 → x = i ⋅ 2 ↔ j ⋅ 2 = i ⋅ 2
4 3 rexbidv ⊢ x = j ⋅ 2 → ∃ i ∈ I x = i ⋅ 2 ↔ ∃ i ∈ I j ⋅ 2 = i ⋅ 2
5 elfzelz ⊢ j ∈ A … B → j ∈ ℤ
6 5 1 eleq2s ⊢ j ∈ I → j ∈ ℤ
7 2z ⊢ 2 ∈ ℤ
8 7 a1i ⊢ j ∈ I → 2 ∈ ℤ
9 6 8 zmulcld ⊢ j ∈ I → j ⋅ 2 ∈ ℤ
10 id ⊢ j ∈ I → j ∈ I
11 oveq1 ⊢ i = j → i ⋅ 2 = j ⋅ 2
12 11 eqeq2d ⊢ i = j → j ⋅ 2 = i ⋅ 2 ↔ j ⋅ 2 = j ⋅ 2
13 12 adantl ⊢ j ∈ I ∧ i = j → j ⋅ 2 = i ⋅ 2 ↔ j ⋅ 2 = j ⋅ 2
14 eqidd ⊢ j ∈ I → j ⋅ 2 = j ⋅ 2
15 10 13 14 rspcedvd ⊢ j ∈ I → ∃ i ∈ I j ⋅ 2 = i ⋅ 2
16 4 9 15 elrabd ⊢ j ∈ I → j ⋅ 2 ∈ x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2
17 2 16 fmpti ⊢ F : I ⟶ x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2
18 oveq1 ⊢ j = y → j ⋅ 2 = y ⋅ 2
19 simpl ⊢ y ∈ I ∧ z ∈ I → y ∈ I
20 ovexd ⊢ y ∈ I ∧ z ∈ I → y ⋅ 2 ∈ V
21 2 18 19 20 fvmptd3 ⊢ y ∈ I ∧ z ∈ I → F ⁡ y = y ⋅ 2
22 oveq1 ⊢ j = z → j ⋅ 2 = z ⋅ 2
23 simpr ⊢ y ∈ I ∧ z ∈ I → z ∈ I
24 ovexd ⊢ y ∈ I ∧ z ∈ I → z ⋅ 2 ∈ V
25 2 22 23 24 fvmptd3 ⊢ y ∈ I ∧ z ∈ I → F ⁡ z = z ⋅ 2
26 21 25 eqeq12d ⊢ y ∈ I ∧ z ∈ I → F ⁡ y = F ⁡ z ↔ y ⋅ 2 = z ⋅ 2
27 elfzelz ⊢ y ∈ A … B → y ∈ ℤ
28 27 1 eleq2s ⊢ y ∈ I → y ∈ ℤ
29 28 zcnd ⊢ y ∈ I → y ∈ ℂ
30 29 adantr ⊢ y ∈ I ∧ z ∈ I → y ∈ ℂ
31 elfzelz ⊢ z ∈ A … B → z ∈ ℤ
32 31 1 eleq2s ⊢ z ∈ I → z ∈ ℤ
33 32 zcnd ⊢ z ∈ I → z ∈ ℂ
34 33 adantl ⊢ y ∈ I ∧ z ∈ I → z ∈ ℂ
35 2cnd ⊢ y ∈ I ∧ z ∈ I → 2 ∈ ℂ
36 2ne0 ⊢ 2 ≠ 0
37 36 a1i ⊢ y ∈ I ∧ z ∈ I → 2 ≠ 0
38 30 34 35 37 mulcan2d ⊢ y ∈ I ∧ z ∈ I → y ⋅ 2 = z ⋅ 2 ↔ y = z
39 38 biimpd ⊢ y ∈ I ∧ z ∈ I → y ⋅ 2 = z ⋅ 2 → y = z
40 26 39 sylbid ⊢ y ∈ I ∧ z ∈ I → F ⁡ y = F ⁡ z → y = z
41 40 rgen2 ⊢ ∀ y ∈ I ∀ z ∈ I F ⁡ y = F ⁡ z → y = z
42 dff13 ⊢ F : I ⟶ 1-1 x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2 ↔ F : I ⟶ x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2 ∧ ∀ y ∈ I ∀ z ∈ I F ⁡ y = F ⁡ z → y = z
43 17 41 42 mpbir2an ⊢ F : I ⟶ 1-1 x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2
44 oveq1 ⊢ j = i → j ⋅ 2 = i ⋅ 2
45 44 eqeq2d ⊢ j = i → x = j ⋅ 2 ↔ x = i ⋅ 2
46 45 cbvrexvw ⊢ ∃ j ∈ I x = j ⋅ 2 ↔ ∃ i ∈ I x = i ⋅ 2
47 elfzelz ⊢ i ∈ A … B → i ∈ ℤ
48 7 a1i ⊢ i ∈ A … B → 2 ∈ ℤ
49 47 48 zmulcld ⊢ i ∈ A … B → i ⋅ 2 ∈ ℤ
50 49 1 eleq2s ⊢ i ∈ I → i ⋅ 2 ∈ ℤ
51 eleq1 ⊢ x = i ⋅ 2 → x ∈ ℤ ↔ i ⋅ 2 ∈ ℤ
52 50 51 syl5ibrcom ⊢ i ∈ I → x = i ⋅ 2 → x ∈ ℤ
53 52 rexlimiv ⊢ ∃ i ∈ I x = i ⋅ 2 → x ∈ ℤ
54 53 pm4.71ri ⊢ ∃ i ∈ I x = i ⋅ 2 ↔ x ∈ ℤ ∧ ∃ i ∈ I x = i ⋅ 2
55 46 54 bitri ⊢ ∃ j ∈ I x = j ⋅ 2 ↔ x ∈ ℤ ∧ ∃ i ∈ I x = i ⋅ 2
56 55 abbii ⊢ x | ∃ j ∈ I x = j ⋅ 2 = x | x ∈ ℤ ∧ ∃ i ∈ I x = i ⋅ 2
57 2 rnmpt ⊢ ran ⁡ F = x | ∃ j ∈ I x = j ⋅ 2
58 df-rab ⊢ x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2 = x | x ∈ ℤ ∧ ∃ i ∈ I x = i ⋅ 2
59 56 57 58 3eqtr4i ⊢ ran ⁡ F = x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2
60 dff1o5 ⊢ F : I ⟶ 1-1 onto x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2 ↔ F : I ⟶ 1-1 x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2 ∧ ran ⁡ F = x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2
61 43 59 60 mpbir2an ⊢ F : I ⟶ 1-1 onto x ∈ ℤ | ∃ i ∈ I x = i ⋅ 2