Metamath Proof Explorer


Theorem binom2sub1

Description: Special case of binom2sub where B = 1 . (Contributed by AV, 2-Aug-2021)

Ref Expression
Assertion binom2sub1 ⊢ A ∈ ℂ → A − 1 2 = A 2 - 2 ⁢ A + 1

Proof

Step Hyp Ref Expression
1 1cnd ⊢ A ∈ ℂ → 1 ∈ ℂ
2 binom2sub ⊢ A ∈ ℂ ∧ 1 ∈ ℂ → A − 1 2 = A 2 - 2 ⁢ A ⋅ 1 + 1 2
3 1 2 mpdan ⊢ A ∈ ℂ → A − 1 2 = A 2 - 2 ⁢ A ⋅ 1 + 1 2
4 mulrid ⊢ A ∈ ℂ → A ⋅ 1 = A
5 4 oveq2d ⊢ A ∈ ℂ → 2 ⁢ A ⋅ 1 = 2 ⁢ A
6 5 oveq2d ⊢ A ∈ ℂ → A 2 − 2 ⁢ A ⋅ 1 = A 2 − 2 ⁢ A
7 sq1 ⊢ 1 2 = 1
8 7 a1i ⊢ A ∈ ℂ → 1 2 = 1
9 6 8 oveq12d ⊢ A ∈ ℂ → A 2 - 2 ⁢ A ⋅ 1 + 1 2 = A 2 - 2 ⁢ A + 1
10 3 9 eqtrd ⊢ A ∈ ℂ → A − 1 2 = A 2 - 2 ⁢ A + 1