Metamath Proof Explorer
Description: Cancellation law for division and multiplication. (Contributed by Mario Carneiro, 27-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
div1d.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
|
|
divcld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
divmuld.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
|
|
divmuld.4 |
⊢ ( 𝜑 → 𝐵 ≠ 0 ) |
|
|
divdiv23d.5 |
⊢ ( 𝜑 → 𝐶 ≠ 0 ) |
|
Assertion |
dmdcand |
⊢ ( 𝜑 → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
div1d.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
2 |
|
divcld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
3 |
|
divmuld.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
4 |
|
divmuld.4 |
⊢ ( 𝜑 → 𝐵 ≠ 0 ) |
5 |
|
divdiv23d.5 |
⊢ ( 𝜑 → 𝐶 ≠ 0 ) |
6 |
|
dmdcan |
⊢ ( ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ 𝐴 ∈ ℂ ) → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) ) |
7 |
2 4 3 5 1 6
|
syl221anc |
⊢ ( 𝜑 → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) ) |