Metamath Proof Explorer

Theorem divcan5

Description: Cancellation of common factor in a ratio. (Contributed by NM, 9-Jan-2006)

		Ref	Expression
	Assertion	divcan5	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to \frac{C A}{C B} = \frac{A}{B}$

Proof

Step	Hyp	Ref	Expression
1		divid	$⊢ (C \in ℂ \land C \neq 0) \to \frac{C}{C} = 1$
2	1	oveq1d	$⊢ (C \in ℂ \land C \neq 0) \to (\frac{C}{C}) (\frac{A}{B}) = 1 (\frac{A}{B})$
3	2	3ad2ant3	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to (\frac{C}{C}) (\frac{A}{B}) = 1 (\frac{A}{B})$
4		simp3l	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to C \in ℂ$
5		simp1	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to A \in ℂ$
6		simp3	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to (C \in ℂ \land C \neq 0)$
7		simp2	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to (B \in ℂ \land B \neq 0)$
8		divmuldiv	$⊢ ((C \in ℂ \land A \in ℂ) \land ((C \in ℂ \land C \neq 0) \land (B \in ℂ \land B \neq 0))) \to (\frac{C}{C}) (\frac{A}{B}) = \frac{C A}{C B}$
9	4 5 6 7 8	syl22anc	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to (\frac{C}{C}) (\frac{A}{B}) = \frac{C A}{C B}$
10		divcl	$⊢ (A \in ℂ \land B \in ℂ \land B \neq 0) \to \frac{A}{B} \in ℂ$
11	10	3expb	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0)) \to \frac{A}{B} \in ℂ$
12	11	mullidd	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0)) \to 1 (\frac{A}{B}) = \frac{A}{B}$
13	12	3adant3	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to 1 (\frac{A}{B}) = \frac{A}{B}$
14	3 9 13	3eqtr3d	$⊢ (A \in ℂ \land (B \in ℂ \land B \neq 0) \land (C \in ℂ \land C \neq 0)) \to \frac{C A}{C B} = \frac{A}{B}$