Metamath Proof Explorer
Description: Move the left term in a product on the LHS to the RHS, deduction form.
(Contributed by David A. Wheeler, 11-Oct-2018)
|
|
Ref |
Expression |
|
Hypotheses |
mvllmuld.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
|
|
mvllmuld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
mvllmuld.3 |
⊢ ( 𝜑 → 𝐴 ≠ 0 ) |
|
|
mvllmuld.4 |
⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) = 𝐶 ) |
|
Assertion |
mvllmuld |
⊢ ( 𝜑 → 𝐵 = ( 𝐶 / 𝐴 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mvllmuld.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
2 |
|
mvllmuld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
3 |
|
mvllmuld.3 |
⊢ ( 𝜑 → 𝐴 ≠ 0 ) |
4 |
|
mvllmuld.4 |
⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) = 𝐶 ) |
5 |
2 1 3
|
divcan4d |
⊢ ( 𝜑 → ( ( 𝐵 · 𝐴 ) / 𝐴 ) = 𝐵 ) |
6 |
1 2
|
mulcomd |
⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) ) |
7 |
6 4
|
eqtr3d |
⊢ ( 𝜑 → ( 𝐵 · 𝐴 ) = 𝐶 ) |
8 |
7
|
oveq1d |
⊢ ( 𝜑 → ( ( 𝐵 · 𝐴 ) / 𝐴 ) = ( 𝐶 / 𝐴 ) ) |
9 |
5 8
|
eqtr3d |
⊢ ( 𝜑 → 𝐵 = ( 𝐶 / 𝐴 ) ) |