Theorem hvm1neg

Description: Convert minus one times a scalar product to the negative of the scalar. (Contributed by NM, 4-Jun-2004) (New usage is discouraged.)

	Ref	Expression
Assertion	hvm1neg	$$⊢\left(A\in ℂ\wedge B\in \mathscr{H}\right)\to -1{\cdot }_{ℎ}\left(A{\cdot }_{ℎ}B\right)=\left(-A\right){\cdot }_{ℎ}B$$

Proof

Step	Hyp	Ref	Expression
1		neg1cn	$$⊢-1\in ℂ$$
2		ax-hvmulass	$$⊢\left(-1\in ℂ\wedge A\in ℂ\wedge B\in \mathscr{H}\right)\to -1A{\cdot }_{ℎ}B=-1{\cdot }_{ℎ}\left(A{\cdot }_{ℎ}B\right)$$
3	1 2	mp3an1	$$⊢\left(A\in ℂ\wedge B\in \mathscr{H}\right)\to -1A{\cdot }_{ℎ}B=-1{\cdot }_{ℎ}\left(A{\cdot }_{ℎ}B\right)$$
4		mulm1	$$⊢A\in ℂ\to -1A=-A$$
5	4	adantr	$$⊢\left(A\in ℂ\wedge B\in \mathscr{H}\right)\to -1A=-A$$
6	5	oveq1d	$$⊢\left(A\in ℂ\wedge B\in \mathscr{H}\right)\to -1A{\cdot }_{ℎ}B=\left(-A\right){\cdot }_{ℎ}B$$
7	3 6	eqtr3d	$$⊢\left(A\in ℂ\wedge B\in \mathscr{H}\right)\to -1{\cdot }_{ℎ}\left(A{\cdot }_{ℎ}B\right)=\left(-A\right){\cdot }_{ℎ}B$$