Metamath Proof Explorer


Theorem divcan2

Description: A cancellation law for division. (Contributed by NM, 3-Feb-2004) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion divcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 eqid ( 𝐴 / 𝐵 ) = ( 𝐴 / 𝐵 )
2 simp1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐴 ∈ ℂ )
3 divcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) ∈ ℂ )
4 3simpc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
5 divmul ( ( 𝐴 ∈ ℂ ∧ ( 𝐴 / 𝐵 ) ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) = ( 𝐴 / 𝐵 ) ↔ ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 ) )
6 2 3 4 5 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) = ( 𝐴 / 𝐵 ) ↔ ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 ) )
7 1 6 mpbii ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )