Metamath Proof Explorer


Theorem addrid

Description: 0 is an additive identity. This used to be one of our complex number axioms, until it was found to be dependent on the others. Based on ideas by Eric Schmidt. (Contributed by Scott Fenton, 3-Jan-2013) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion addrid ⊢ A ∈ ℂ → A + 0 = A

Proof

Step Hyp Ref Expression
1 1re ⊢ 1 ∈ ℝ
2 ax-rnegex ⊢ 1 ∈ ℝ → ∃ c ∈ ℝ 1 + c = 0
3 ax-1ne0 ⊢ 1 ≠ 0
4 oveq2 ⊢ c = 0 → 1 + c = 1 + 0
5 4 eqeq1d ⊢ c = 0 → 1 + c = 0 ↔ 1 + 0 = 0
6 5 biimpcd ⊢ 1 + c = 0 → c = 0 → 1 + 0 = 0
7 oveq2 ⊢ 1 + 0 = 0 → i ⁢ i ⁢ i ⁢ i ⁢ 1 + 0 = i ⁢ i ⁢ i ⁢ i ⋅ 0
8 ax-icn ⊢ i ∈ ℂ
9 8 8 mulcli ⊢ i ⁢ i ∈ ℂ
10 9 9 mulcli ⊢ i ⁢ i ⁢ i ⁢ i ∈ ℂ
11 ax-1cn ⊢ 1 ∈ ℂ
12 0cn ⊢ 0 ∈ ℂ
13 10 11 12 adddii ⊢ i ⁢ i ⁢ i ⁢ i ⁢ 1 + 0 = i ⁢ i ⁢ i ⁢ i ⋅ 1 + i ⁢ i ⁢ i ⁢ i ⋅ 0
14 10 mulridi ⊢ i ⁢ i ⁢ i ⁢ i ⋅ 1 = i ⁢ i ⁢ i ⁢ i
15 mul01 ⊢ i ⁢ i ⁢ i ⁢ i ∈ ℂ → i ⁢ i ⁢ i ⁢ i ⋅ 0 = 0
16 10 15 ax-mp ⊢ i ⁢ i ⁢ i ⁢ i ⋅ 0 = 0
17 ax-i2m1 ⊢ i ⁢ i + 1 = 0
18 16 17 eqtr4i ⊢ i ⁢ i ⁢ i ⁢ i ⋅ 0 = i ⁢ i + 1
19 14 18 oveq12i ⊢ i ⁢ i ⁢ i ⁢ i ⋅ 1 + i ⁢ i ⁢ i ⁢ i ⋅ 0 = i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1
20 13 19 eqtri ⊢ i ⁢ i ⁢ i ⁢ i ⁢ 1 + 0 = i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1
21 20 16 eqeq12i ⊢ i ⁢ i ⁢ i ⁢ i ⁢ 1 + 0 = i ⁢ i ⁢ i ⁢ i ⋅ 0 ↔ i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1 = 0
22 10 9 11 addassi ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1 = i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1
23 9 mulridi ⊢ i ⁢ i ⋅ 1 = i ⁢ i
24 23 oveq2i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i ⋅ 1 = i ⁢ i ⁢ i ⁢ i + i ⁢ i
25 9 9 11 adddii ⊢ i ⁢ i ⁢ i ⁢ i + 1 = i ⁢ i ⁢ i ⁢ i + i ⁢ i ⋅ 1
26 17 oveq2i ⊢ i ⁢ i ⁢ i ⁢ i + 1 = i ⁢ i ⋅ 0
27 mul01 ⊢ i ⁢ i ∈ ℂ → i ⁢ i ⋅ 0 = 0
28 9 27 ax-mp ⊢ i ⁢ i ⋅ 0 = 0
29 26 28 eqtri ⊢ i ⁢ i ⁢ i ⁢ i + 1 = 0
30 25 29 eqtr3i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i ⋅ 1 = 0
31 24 30 eqtr3i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i = 0
32 31 oveq1i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1 = 0 + 1
33 22 32 eqtr3i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1 = 0 + 1
34 00id ⊢ 0 + 0 = 0
35 34 eqcomi ⊢ 0 = 0 + 0
36 33 35 eqeq12i ⊢ i ⁢ i ⁢ i ⁢ i + i ⁢ i + 1 = 0 ↔ 0 + 1 = 0 + 0
37 0re ⊢ 0 ∈ ℝ
38 readdcan ⊢ 1 ∈ ℝ ∧ 0 ∈ ℝ ∧ 0 ∈ ℝ → 0 + 1 = 0 + 0 ↔ 1 = 0
39 1 37 37 38 mp3an ⊢ 0 + 1 = 0 + 0 ↔ 1 = 0
40 21 36 39 3bitri ⊢ i ⁢ i ⁢ i ⁢ i ⁢ 1 + 0 = i ⁢ i ⁢ i ⁢ i ⋅ 0 ↔ 1 = 0
41 7 40 sylib ⊢ 1 + 0 = 0 → 1 = 0
42 6 41 syl6 ⊢ 1 + c = 0 → c = 0 → 1 = 0
43 42 necon3d ⊢ 1 + c = 0 → 1 ≠ 0 → c ≠ 0
44 3 43 mpi ⊢ 1 + c = 0 → c ≠ 0
45 ax-rrecex ⊢ c ∈ ℝ ∧ c ≠ 0 → ∃ x ∈ ℝ c ⁢ x = 1
46 44 45 sylan2 ⊢ c ∈ ℝ ∧ 1 + c = 0 → ∃ x ∈ ℝ c ⁢ x = 1
47 simpr ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ∈ ℂ
48 simplrl ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → x ∈ ℝ
49 48 recnd ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → x ∈ ℂ
50 47 49 mulcld ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ∈ ℂ
51 simplll ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c ∈ ℝ
52 51 recnd ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c ∈ ℂ
53 12 a1i ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 0 ∈ ℂ
54 50 52 53 adddid ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c + 0 = A ⁢ x ⁢ c + A ⁢ x ⋅ 0
55 11 a1i ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 ∈ ℂ
56 55 52 53 addassd ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c + 0 = 1 + c + 0
57 simpllr ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c = 0
58 57 oveq1d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c + 0 = 0 + 0
59 56 58 eqtr3d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c + 0 = 0 + 0
60 34 59 57 3eqtr4a ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c + 0 = 1 + c
61 37 a1i ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 0 ∈ ℝ
62 51 61 readdcld ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c + 0 ∈ ℝ
63 1 a1i ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 ∈ ℝ
64 readdcan ⊢ c + 0 ∈ ℝ ∧ c ∈ ℝ ∧ 1 ∈ ℝ → 1 + c + 0 = 1 + c ↔ c + 0 = c
65 62 51 63 64 syl3anc ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 + c + 0 = 1 + c ↔ c + 0 = c
66 60 65 mpbid ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c + 0 = c
67 66 oveq2d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c + 0 = A ⁢ x ⁢ c
68 54 67 eqtr3d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c + A ⁢ x ⋅ 0 = A ⁢ x ⁢ c
69 mul31 ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ c ∈ ℂ → A ⁢ x ⁢ c = c ⁢ x ⁢ A
70 47 49 52 69 syl3anc ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c = c ⁢ x ⁢ A
71 simplrr ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c ⁢ x = 1
72 71 oveq1d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → c ⁢ x ⁢ A = 1 ⁢ A
73 47 mullidd ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → 1 ⁢ A = A
74 70 72 73 3eqtrd ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c = A
75 mul01 ⊢ A ⁢ x ∈ ℂ → A ⁢ x ⋅ 0 = 0
76 50 75 syl ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⋅ 0 = 0
77 74 76 oveq12d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A ⁢ x ⁢ c + A ⁢ x ⋅ 0 = A + 0
78 68 77 74 3eqtr3d ⊢ c ∈ ℝ ∧ 1 + c = 0 ∧ x ∈ ℝ ∧ c ⁢ x = 1 ∧ A ∈ ℂ → A + 0 = A
79 78 exp42 ⊢ c ∈ ℝ ∧ 1 + c = 0 → x ∈ ℝ → c ⁢ x = 1 → A ∈ ℂ → A + 0 = A
80 79 rexlimdv ⊢ c ∈ ℝ ∧ 1 + c = 0 → ∃ x ∈ ℝ c ⁢ x = 1 → A ∈ ℂ → A + 0 = A
81 46 80 mpd ⊢ c ∈ ℝ ∧ 1 + c = 0 → A ∈ ℂ → A + 0 = A
82 81 rexlimiva ⊢ ∃ c ∈ ℝ 1 + c = 0 → A ∈ ℂ → A + 0 = A
83 1 2 82 mp2b ⊢ A ∈ ℂ → A + 0 = A