Metamath Proof Explorer


Theorem sn-it0e0

Description: Proof of it0e0 without ax-mulcom . Informally, a real number times 0 is 0, and E. r e. RR r = _i x. s by ax-cnre and renegid2 . (Contributed by SN, 30-Apr-2024)

Ref Expression
Assertion sn-it0e0 ⊢ i ⋅ 0 = 0

Proof

Step Hyp Ref Expression
1 0cn ⊢ 0 ∈ ℂ
2 cnre ⊢ 0 ∈ ℂ → ∃ a ∈ ℝ ∃ b ∈ ℝ 0 = a + i ⁢ b
3 oveq2 ⊢ 0 = a + i ⁢ b → 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b
4 ax-icn ⊢ i ∈ ℂ
5 4 a1i ⊢ b ∈ ℝ → i ∈ ℂ
6 recn ⊢ b ∈ ℝ → b ∈ ℂ
7 0cnd ⊢ b ∈ ℝ → 0 ∈ ℂ
8 5 6 7 mulassd ⊢ b ∈ ℝ → i ⁢ b ⋅ 0 = i ⁢ b ⋅ 0
9 remul01 ⊢ b ∈ ℝ → b ⋅ 0 = 0
10 9 oveq2d ⊢ b ∈ ℝ → i ⁢ b ⋅ 0 = i ⋅ 0
11 8 10 eqtrd ⊢ b ∈ ℝ → i ⁢ b ⋅ 0 = i ⋅ 0
12 11 ad2antlr ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → i ⁢ b ⋅ 0 = i ⋅ 0
13 rernegcl ⊢ a ∈ ℝ → 0 - ℝ a ∈ ℝ
14 13 recnd ⊢ a ∈ ℝ → 0 - ℝ a ∈ ℂ
15 14 adantr ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a ∈ ℂ
16 recn ⊢ a ∈ ℝ → a ∈ ℂ
17 16 adantr ⊢ a ∈ ℝ ∧ b ∈ ℝ → a ∈ ℂ
18 5 6 mulcld ⊢ b ∈ ℝ → i ⁢ b ∈ ℂ
19 18 adantl ⊢ a ∈ ℝ ∧ b ∈ ℝ → i ⁢ b ∈ ℂ
20 15 17 19 addassd ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a + a + i ⁢ b = 0 - ℝ a + a + i ⁢ b
21 renegid2 ⊢ a ∈ ℝ → 0 - ℝ a + a = 0
22 21 oveq1d ⊢ a ∈ ℝ → 0 - ℝ a + a + i ⁢ b = 0 + i ⁢ b
23 sn-addlid ⊢ i ⁢ b ∈ ℂ → 0 + i ⁢ b = i ⁢ b
24 18 23 syl ⊢ b ∈ ℝ → 0 + i ⁢ b = i ⁢ b
25 22 24 sylan9eq ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a + a + i ⁢ b = i ⁢ b
26 20 25 eqtr3d ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a + a + i ⁢ b = i ⁢ b
27 26 eqeq2d ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b ↔ 0 - ℝ a + 0 = i ⁢ b
28 27 biimpa ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → 0 - ℝ a + 0 = i ⁢ b
29 28 oveq1d ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → 0 - ℝ a + 0 ⋅ 0 = i ⁢ b ⋅ 0
30 elre0re ⊢ a ∈ ℝ → 0 ∈ ℝ
31 13 30 readdcld ⊢ a ∈ ℝ → 0 - ℝ a + 0 ∈ ℝ
32 31 ad2antrr ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → 0 - ℝ a + 0 ∈ ℝ
33 remul01 ⊢ 0 - ℝ a + 0 ∈ ℝ → 0 - ℝ a + 0 ⋅ 0 = 0
34 32 33 syl ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → 0 - ℝ a + 0 ⋅ 0 = 0
35 29 34 eqtr3d ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → i ⁢ b ⋅ 0 = 0
36 12 35 eqtr3d ⊢ a ∈ ℝ ∧ b ∈ ℝ ∧ 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → i ⋅ 0 = 0
37 36 ex ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 - ℝ a + 0 = 0 - ℝ a + a + i ⁢ b → i ⋅ 0 = 0
38 3 37 syl5 ⊢ a ∈ ℝ ∧ b ∈ ℝ → 0 = a + i ⁢ b → i ⋅ 0 = 0
39 38 rexlimivv ⊢ ∃ a ∈ ℝ ∃ b ∈ ℝ 0 = a + i ⁢ b → i ⋅ 0 = 0
40 1 2 39 mp2b ⊢ i ⋅ 0 = 0