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Metamath Proof Explorer

Theorem divcan8d

Description: A cancellation law for division. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses divcan8d.a ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
divcan8d.b ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
divcan8d.a0 ⊢ ( 𝜑 → 𝐴 ≠ 0 )
divcan8d.b0 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
Assertion divcan8d ( 𝜑 → ( 𝐵 / ( 𝐴 · 𝐵 ) ) = ( 1 / 𝐴 ) )

Proof

Step Hyp Ref Expression
1 divcan8d.a ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 divcan8d.b ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 divcan8d.a0 ⊢ ( 𝜑 → 𝐴 ≠ 0 )
4 divcan8d.b0 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
5 1 2 mulcld ⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) ∈ ℂ )
6 1 2 3 4 mulne0d ⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) ≠ 0 )
7 1 2 6 mulne0bbd ⊢ ( 𝜑 → 𝐵 ≠ 0 )
8 2 5 2 6 7 divcan7d ⊢ ( 𝜑 → ( ( 𝐵 / 𝐵 ) / ( ( 𝐴 · 𝐵 ) / 𝐵 ) ) = ( 𝐵 / ( 𝐴 · 𝐵 ) ) )
9 8 eqcomd ⊢ ( 𝜑 → ( 𝐵 / ( 𝐴 · 𝐵 ) ) = ( ( 𝐵 / 𝐵 ) / ( ( 𝐴 · 𝐵 ) / 𝐵 ) ) )
10 2 4 dividd ⊢ ( 𝜑 → ( 𝐵 / 𝐵 ) = 1 )
11 1 2 4 divcan4d ⊢ ( 𝜑 → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )
12 10 11 oveq12d ⊢ ( 𝜑 → ( ( 𝐵 / 𝐵 ) / ( ( 𝐴 · 𝐵 ) / 𝐵 ) ) = ( 1 / 𝐴 ) )
13 eqidd ⊢ ( 𝜑 → ( 1 / 𝐴 ) = ( 1 / 𝐴 ) )
14 9 12 13 3eqtrd ⊢ ( 𝜑 → ( 𝐵 / ( 𝐴 · 𝐵 ) ) = ( 1 / 𝐴 ) )