Metamath Proof Explorer

Theorem ddcan

Description: Cancellation in a double division. (Contributed by Mario Carneiro, 26-Apr-2015)

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

Proof

Step Hyp Ref Expression
1 simpll ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → 𝐴 ∈ ℂ )
2 simprl ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → 𝐵 ∈ ℂ )
3 simprr ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → 𝐵 ≠ 0 )
4 divcan1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 )
5 1 2 3 4 syl3anc ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 )
6 divcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) ∈ ℂ )
7 1 2 3 6 syl3anc ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( 𝐴 / 𝐵 ) ∈ ℂ )
8 divne0 ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( 𝐴 / 𝐵 ) ≠ 0 )
9 divmul ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( ( 𝐴 / 𝐵 ) ∈ ℂ ∧ ( 𝐴 / 𝐵 ) ≠ 0 ) ) → ( ( 𝐴 / ( 𝐴 / 𝐵 ) ) = 𝐵 ↔ ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 ) )
10 1 2 7 8 9 syl112anc ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 𝐴 / ( 𝐴 / 𝐵 ) ) = 𝐵 ↔ ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 ) )
11 5 10 mpbird ⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( 𝐴 / ( 𝐴 / 𝐵 ) ) = 𝐵 )