Metamath Proof Explorer


Theorem remullid

Description: Commuted version of ax-1rid without ax-mulcom . (Contributed by SN, 5-Feb-2024)

Ref Expression
Assertion remullid ⊢ A ∈ ℝ → 1 ⁢ A = A

Proof

Step Hyp Ref Expression
1 df-ne ⊢ A ≠ 0 ↔ ¬ A = 0
2 ax-rrecex ⊢ A ∈ ℝ ∧ A ≠ 0 → ∃ x ∈ ℝ A ⁢ x = 1
3 simpll ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ∈ ℝ
4 3 recnd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ∈ ℂ
5 simprl ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → x ∈ ℝ
6 5 recnd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → x ∈ ℂ
7 4 6 4 mulassd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⁢ x ⁢ A = A ⁢ x ⁢ A
8 simprr ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⁢ x = 1
9 8 oveq1d ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⁢ x ⁢ A = 1 ⁢ A
10 3 5 8 remulinvcom ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → x ⁢ A = 1
11 10 oveq2d ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⁢ x ⁢ A = A ⋅ 1
12 ax-1rid ⊢ A ∈ ℝ → A ⋅ 1 = A
13 3 12 syl ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⋅ 1 = A
14 11 13 eqtrd ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → A ⁢ x ⁢ A = A
15 7 9 14 3eqtr3d ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ x ∈ ℝ ∧ A ⁢ x = 1 → 1 ⁢ A = A
16 2 15 rexlimddv ⊢ A ∈ ℝ ∧ A ≠ 0 → 1 ⁢ A = A
17 16 ex ⊢ A ∈ ℝ → A ≠ 0 → 1 ⁢ A = A
18 1 17 biimtrrid ⊢ A ∈ ℝ → ¬ A = 0 → 1 ⁢ A = A
19 1re ⊢ 1 ∈ ℝ
20 remul01 ⊢ 1 ∈ ℝ → 1 ⋅ 0 = 0
21 19 20 mp1i ⊢ A = 0 → 1 ⋅ 0 = 0
22 oveq2 ⊢ A = 0 → 1 ⁢ A = 1 ⋅ 0
23 id ⊢ A = 0 → A = 0
24 21 22 23 3eqtr4d ⊢ A = 0 → 1 ⁢ A = A
25 18 24 pm2.61d2 ⊢ A ∈ ℝ → 1 ⁢ A = A