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Metamath Proof Explorer

Theorem divcan2zi

Description: A cancellation law for division. (Contributed by NM, 10-Aug-1999)

Ref Expression
Hypotheses divclz.1 ⊢ 𝐴 ∈ ℂ
divclz.2 ⊢ 𝐵 ∈ ℂ
Assertion divcan2zi ( 𝐵 ≠ 0 → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 divclz.1 ⊢ 𝐴 ∈ ℂ
2 divclz.2 ⊢ 𝐵 ∈ ℂ
3 divcan2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )
4 1 2 3 mp3an12 ⊢ ( 𝐵 ≠ 0 → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )