Metamath Proof Explorer


Theorem divval

Description: Value of division: if A and B are complex numbers with B nonzero, then ( A / B ) is the (unique) complex number such that ( B x. x ) = A . (Contributed by NM, 8-May-1999) (Revised by Mario Carneiro, 17-Feb-2014)

Ref Expression
Assertion divval ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 eldifsn ( 𝐵 ∈ ( ℂ ∖ { 0 } ) ↔ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
2 eqeq2 ( 𝑧 = 𝐴 → ( ( 𝑦 · 𝑥 ) = 𝑧 ↔ ( 𝑦 · 𝑥 ) = 𝐴 ) )
3 2 riotabidv ( 𝑧 = 𝐴 → ( 𝑥 ∈ ℂ ( 𝑦 · 𝑥 ) = 𝑧 ) = ( 𝑥 ∈ ℂ ( 𝑦 · 𝑥 ) = 𝐴 ) )
4 oveq1 ( 𝑦 = 𝐵 → ( 𝑦 · 𝑥 ) = ( 𝐵 · 𝑥 ) )
5 4 eqeq1d ( 𝑦 = 𝐵 → ( ( 𝑦 · 𝑥 ) = 𝐴 ↔ ( 𝐵 · 𝑥 ) = 𝐴 ) )
6 5 riotabidv ( 𝑦 = 𝐵 → ( 𝑥 ∈ ℂ ( 𝑦 · 𝑥 ) = 𝐴 ) = ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) )
7 df-div / = ( 𝑧 ∈ ℂ , 𝑦 ∈ ( ℂ ∖ { 0 } ) ↦ ( 𝑥 ∈ ℂ ( 𝑦 · 𝑥 ) = 𝑧 ) )
8 riotaex ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) ∈ V
9 3 6 7 8 ovmpo ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ( ℂ ∖ { 0 } ) ) → ( 𝐴 / 𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) )
10 1 9 sylan2br ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( 𝐴 / 𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) )
11 10 3impb ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 · 𝑥 ) = 𝐴 ) )