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Metamath Proof Explorer

Theorem redivrec2d

Description: Relationship between division and reciprocal. (Contributed by SN, 9-Apr-2026)

Ref Expression
Hypotheses redivrec2d.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
redivrec2d.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
redivrec2d.z ⊢ ( 𝜑 → 𝐵 ≠ 0 )
Assertion redivrec2d ( 𝜑 → ( 𝐴 /ℝ 𝐵 ) = ( ( 1 /ℝ 𝐵 ) · 𝐴 ) )

Proof

Step Hyp Ref Expression
1 redivrec2d.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2 redivrec2d.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
3 redivrec2d.z ⊢ ( 𝜑 → 𝐵 ≠ 0 )
4 2 3 rerecidd ⊢ ( 𝜑 → ( 𝐵 · ( 1 /ℝ 𝐵 ) ) = 1 )
5 4 oveq1d ⊢ ( 𝜑 → ( ( 𝐵 · ( 1 /ℝ 𝐵 ) ) · 𝐴 ) = ( 1 · 𝐴 ) )
6 2 recnd ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
7 2 3 sn-rereccld ⊢ ( 𝜑 → ( 1 /ℝ 𝐵 ) ∈ ℝ )
8 7 recnd ⊢ ( 𝜑 → ( 1 /ℝ 𝐵 ) ∈ ℂ )
9 1 recnd ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
10 6 8 9 mulassd ⊢ ( 𝜑 → ( ( 𝐵 · ( 1 /ℝ 𝐵 ) ) · 𝐴 ) = ( 𝐵 · ( ( 1 /ℝ 𝐵 ) · 𝐴 ) ) )
11 remullid ⊢ ( 𝐴 ∈ ℝ → ( 1 · 𝐴 ) = 𝐴 )
12 1 11 syl ⊢ ( 𝜑 → ( 1 · 𝐴 ) = 𝐴 )
13 5 10 12 3eqtr3d ⊢ ( 𝜑 → ( 𝐵 · ( ( 1 /ℝ 𝐵 ) · 𝐴 ) ) = 𝐴 )
14 7 1 remulcld ⊢ ( 𝜑 → ( ( 1 /ℝ 𝐵 ) · 𝐴 ) ∈ ℝ )
15 1 14 2 3 redivmuld ⊢ ( 𝜑 → ( ( 𝐴 /ℝ 𝐵 ) = ( ( 1 /ℝ 𝐵 ) · 𝐴 ) ↔ ( 𝐵 · ( ( 1 /ℝ 𝐵 ) · 𝐴 ) ) = 𝐴 ) )
16 13 15 mpbird ⊢ ( 𝜑 → ( 𝐴 /ℝ 𝐵 ) = ( ( 1 /ℝ 𝐵 ) · 𝐴 ) )