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Metamath Proof Explorer

Theorem divass

Description: An associative law for division. (Contributed by NM, 2-Aug-2004)

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

Proof

Step Hyp Ref Expression
1 reccl ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 1 / 𝐶 ) ∈ ℂ )
2 mulass ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 1 / 𝐶 ) ∈ ℂ ) → ( ( 𝐴 · 𝐵 ) · ( 1 / 𝐶 ) ) = ( 𝐴 · ( 𝐵 · ( 1 / 𝐶 ) ) ) )
3 1 2 syl3an3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 𝐵 ) · ( 1 / 𝐶 ) ) = ( 𝐴 · ( 𝐵 · ( 1 / 𝐶 ) ) ) )
4 mulcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 · 𝐵 ) ∈ ℂ )
5 4 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 · 𝐵 ) ∈ ℂ )
6 simp3l ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐶 ∈ ℂ )
7 simp3r ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐶 ≠ 0 )
8 divrec ⊢ ( ( ( 𝐴 · 𝐵 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( ( 𝐴 · 𝐵 ) · ( 1 / 𝐶 ) ) )
9 5 6 7 8 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( ( 𝐴 · 𝐵 ) · ( 1 / 𝐶 ) ) )
10 simp2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → 𝐵 ∈ ℂ )
11 divrec ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐵 / 𝐶 ) = ( 𝐵 · ( 1 / 𝐶 ) ) )
12 10 6 7 11 syl3anc ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) = ( 𝐵 · ( 1 / 𝐶 ) ) )
13 12 oveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 · ( 𝐵 / 𝐶 ) ) = ( 𝐴 · ( 𝐵 · ( 1 / 𝐶 ) ) ) )
14 3 9 13 3eqtr4d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )