Theorem divcan6d

Description: Cancellation of inverted fractions. (Contributed by Mario Carneiro, 27-May-2016)

	Ref	Expression
Hypotheses	div1d.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
	divcld.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
	divne0d.3	⊢ ( 𝜑 → 𝐴 ≠ 0 )
	divne0d.4	⊢ ( 𝜑 → 𝐵 ≠ 0 )
Assertion	divcan6d	⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐴 ) ) = 1 )

Proof

Step	Hyp	Ref	Expression
1		div1d.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2		divcld.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3		divne0d.3	⊢ ( 𝜑 → 𝐴 ≠ 0 )
4		divne0d.4	⊢ ( 𝜑 → 𝐵 ≠ 0 )
5		divcan6	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐴 ) ) = 1 )
6	1 3 2 4 5	syl22anc	⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐴 ) ) = 1 )