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Metamath Proof Explorer

Theorem caov12d

Description: Rearrange arguments in a commutative, associative operation. (Contributed by NM, 26-Aug-1995) (Revised by Mario Carneiro, 30-Dec-2014)

		Ref	Expression
	Hypotheses	caovd.1	$⊢ φ \to A \in S$
		caovd.2	$⊢ φ \to B \in S$
		caovd.3	$⊢ φ \to C \in S$
		caovd.com	$⊢ (φ \land (x \in S \land y \in S)) \to x F y = y F x$
		caovd.ass	$⊢ (φ \land (x \in S \land y \in S \land z \in S)) \to (x F y) F z = x F (y F z)$
	Assertion	caov12d	$⊢ φ \to A F (B F C) = B F (A F C)$

Proof

Step	Hyp	Ref	Expression
1		caovd.1	$⊢ φ \to A \in S$
2		caovd.2	$⊢ φ \to B \in S$
3		caovd.3	$⊢ φ \to C \in S$
4		caovd.com	$⊢ (φ \land (x \in S \land y \in S)) \to x F y = y F x$
5		caovd.ass	$⊢ (φ \land (x \in S \land y \in S \land z \in S)) \to (x F y) F z = x F (y F z)$
6	4 1 2	caovcomd	$⊢ φ \to A F B = B F A$
7	6	oveq1d	$⊢ φ \to (A F B) F C = (B F A) F C$
8	5 1 2 3	caovassd	$⊢ φ \to (A F B) F C = A F (B F C)$
9	5 2 1 3	caovassd	$⊢ φ \to (B F A) F C = B F (A F C)$
10	7 8 9	3eqtr3d	$⊢ φ \to A F (B F C) = B F (A F C)$