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Metamath Proof Explorer

Theorem divcan3

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

Ref Expression
Assertion divcan3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 eqid ⊢ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 )
2 simp2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐵 ∈ ℂ )
3 simp1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐴 ∈ ℂ )
4 2 3 mulcld ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · 𝐴 ) ∈ ℂ )
5 3simpc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
6 divmul ⊢ ( ( ( 𝐵 · 𝐴 ) ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 ↔ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 ) ) )
7 4 3 5 6 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 ↔ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 ) ) )
8 1 7 mpbiri ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )