Theorem mulnegs2d

Description: Product with negative is negative of product. Part of theorem 7 of Conway p. 19. (Contributed by Scott Fenton, 10-Mar-2025)

	Ref	Expression
Hypotheses	mulnegs1d.1	$$⊢\phi \to A\in \mathrm{No}$$
	mulnegs1d.2	$$⊢\phi \to B\in \mathrm{No}$$
Assertion	mulnegs2d	$$⊢\phi \to A{\cdot }_s{+}_s\left(B\right)={+}_s\left(A{\cdot }_{s}B\right)$$

Proof

Step	Hyp	Ref	Expression
1		mulnegs1d.1	$$⊢\phi \to A\in \mathrm{No}$$
2		mulnegs1d.2	$$⊢\phi \to B\in \mathrm{No}$$
3	2 1	mulnegs1d	$$⊢\phi \to {+}_s\left(B\right){\cdot }_{s}A={+}_s\left(B{\cdot }_{s}A\right)$$
4	2	negscld	$$⊢\phi \to {+}_s\left(B\right)\in \mathrm{No}$$
5	1 4	mulscomd	$$⊢\phi \to A{\cdot }_s{+}_s\left(B\right)={+}_s\left(B\right){\cdot }_{s}A$$
6	1 2	mulscomd	$$⊢\phi \to A{\cdot }_{s}B=B{\cdot }_{s}A$$
7	6	fveq2d	$$⊢\phi \to {+}_s\left(A{\cdot }_{s}B\right)={+}_s\left(B{\cdot }_{s}A\right)$$
8	3 5 7	3eqtr4d	$$⊢\phi \to A{\cdot }_s{+}_s\left(B\right)={+}_s\left(A{\cdot }_{s}B\right)$$