Metamath Proof Explorer


Theorem sn-addrid

Description: addrid without ax-mulcom . (Contributed by SN, 5-May-2024)

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

Proof

Step Hyp Ref Expression
1 sn-negex2 ⊢ A ∈ ℂ → ∃ x ∈ ℂ x + A = 0
2 simprr ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A = 0
3 2 oveq1d ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A + 0 = 0 + 0
4 sn-00id ⊢ 0 + 0 = 0
5 3 4 eqtrdi ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A + 0 = 0
6 simprl ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x ∈ ℂ
7 simpl ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → A ∈ ℂ
8 0cnd ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → 0 ∈ ℂ
9 6 7 8 addassd ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A + 0 = x + A + 0
10 2 5 9 3eqtr2rd ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A + 0 = x + A
11 7 8 addcld ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → A + 0 ∈ ℂ
12 6 11 7 sn-addcand ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → x + A + 0 = x + A ↔ A + 0 = A
13 10 12 mpbid ⊢ A ∈ ℂ ∧ x ∈ ℂ ∧ x + A = 0 → A + 0 = A
14 1 13 rexlimddv ⊢ A ∈ ℂ → A + 0 = A