Theorem mulnegs1d

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	mulnegs1d	$$⊢\phi \to {+}_s\left(A\right){\cdot }_{s}B={+}_s\left(A{\cdot }_{s}B\right)$$

Proof

Step	Hyp	Ref	Expression
1		mulnegs1d.1	$$⊢\phi \to A\in \mathrm{No}$$
2		mulnegs1d.2	$$⊢\phi \to B\in \mathrm{No}$$
3	1	negsidd	$$⊢\phi \to A{+}_s{+}_s\left(A\right)=0_s$$
4	3	oveq1d	$$⊢\phi \to \left(A{+}_s{+}_s\left(A\right)\right){\cdot }_{s}B=0_s{\cdot }_{s}B$$
5	1	negscld	$$⊢\phi \to {+}_s\left(A\right)\in \mathrm{No}$$
6	1 5 2	addsdird	$$⊢\phi \to \left(A{+}_s{+}_s\left(A\right)\right){\cdot }_{s}B=\left(A{\cdot }_{s}B\right){+}_s\left({+}_s\left(A\right){\cdot }_{s}B\right)$$
7		muls02	$$⊢B\in \mathrm{No}\to 0_s{\cdot }_{s}B=0_s$$
8	2 7	syl	$$⊢\phi \to 0_s{\cdot }_{s}B=0_s$$
9	4 6 8	3eqtr3d	$$⊢\phi \to \left(A{\cdot }_{s}B\right){+}_s\left({+}_s\left(A\right){\cdot }_{s}B\right)=0_s$$
10	1 2	mulscld	$$⊢\phi \to A{\cdot }_{s}B\in \mathrm{No}$$
11	10	negsidd	$$⊢\phi \to \left(A{\cdot }_{s}B\right){+}_s{+}_s\left(A{\cdot }_{s}B\right)=0_s$$
12	9 11	eqtr4d	$$⊢\phi \to \left(A{\cdot }_{s}B\right){+}_s\left({+}_s\left(A\right){\cdot }_{s}B\right)=\left(A{\cdot }_{s}B\right){+}_s{+}_s\left(A{\cdot }_{s}B\right)$$
13	5 2	mulscld	$$⊢\phi \to {+}_s\left(A\right){\cdot }_{s}B\in \mathrm{No}$$
14	10	negscld	$$⊢\phi \to {+}_s\left(A{\cdot }_{s}B\right)\in \mathrm{No}$$
15	13 14 10	addscan1d	$$⊢\phi \to \left(\left(A{\cdot }_{s}B\right){+}_s\left({+}_s\left(A\right){\cdot }_{s}B\right)=\left(A{\cdot }_{s}B\right){+}_s{+}_s\left(A{\cdot }_{s}B\right)\leftrightarrow {+}_s\left(A\right){\cdot }_{s}B={+}_s\left(A{\cdot }_{s}B\right)\right)$$
16	12 15	mpbid	$$⊢\phi \to {+}_s\left(A\right){\cdot }_{s}B={+}_s\left(A{\cdot }_{s}B\right)$$