Metamath Proof Explorer


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 ⊢ ( 𝜑 → 𝐴 ∈ No )
mulnegs1d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
Assertion mulnegs2d ( 𝜑 → ( 𝐴 ·s ( -us ‘ 𝐵 ) ) = ( -us ‘ ( 𝐴 ·s 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 mulnegs1d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 mulnegs1d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 2 1 mulnegs1d ⊢ ( 𝜑 → ( ( -us ‘ 𝐵 ) ·s 𝐴 ) = ( -us ‘ ( 𝐵 ·s 𝐴 ) ) )
4 2 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐵 ) ∈ No )
5 1 4 mulscomd ⊢ ( 𝜑 → ( 𝐴 ·s ( -us ‘ 𝐵 ) ) = ( ( -us ‘ 𝐵 ) ·s 𝐴 ) )
6 1 2 mulscomd ⊢ ( 𝜑 → ( 𝐴 ·s 𝐵 ) = ( 𝐵 ·s 𝐴 ) )
7 6 fveq2d ⊢ ( 𝜑 → ( -us ‘ ( 𝐴 ·s 𝐵 ) ) = ( -us ‘ ( 𝐵 ·s 𝐴 ) ) )
8 3 5 7 3eqtr4d ⊢ ( 𝜑 → ( 𝐴 ·s ( -us ‘ 𝐵 ) ) = ( -us ‘ ( 𝐴 ·s 𝐵 ) ) )