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Metamath Proof Explorer

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 )