Metamath Proof Explorer

Theorem dmdcan

Description: Cancellation law for division and multiplication. (Contributed by Scott Fenton, 7-Jun-2013) (Proof shortened by Fan Zheng, 3-Jul-2016)

		Ref	Expression
	Assertion	dmdcan	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to (\frac{A}{B}) (\frac{C}{A}) = \frac{C}{B}$

Proof

Step	Hyp	Ref	Expression
1		simp1l	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to A \in ℂ$
2		simp3	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to C \in ℂ$
3		simp1r	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to A \neq 0$
4		divcl	$⊢ (C \in ℂ \land A \in ℂ \land A \neq 0) \to \frac{C}{A} \in ℂ$
5	2 1 3 4	syl3anc	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to \frac{C}{A} \in ℂ$
6		simp2l	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to B \in ℂ$
7		simp2r	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to B \neq 0$
8		div23	$⊢ (A \in ℂ \land \frac{C}{A} \in ℂ \land (B \in ℂ \land B \neq 0)) \to \frac{A (\frac{C}{A})}{B} = (\frac{A}{B}) (\frac{C}{A})$
9	1 5 6 7 8	syl112anc	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to \frac{A (\frac{C}{A})}{B} = (\frac{A}{B}) (\frac{C}{A})$
10		divcan2	$⊢ (C \in ℂ \land A \in ℂ \land A \neq 0) \to A (\frac{C}{A}) = C$
11	2 1 3 10	syl3anc	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to A (\frac{C}{A}) = C$
12	11	oveq1d	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to \frac{A (\frac{C}{A})}{B} = \frac{C}{B}$
13	9 12	eqtr3d	$⊢ ((A \in ℂ \land A \neq 0) \land (B \in ℂ \land B \neq 0) \land C \in ℂ) \to (\frac{A}{B}) (\frac{C}{A}) = \frac{C}{B}$