Metamath Proof Explorer


Theorem addmuldir1d

Description: Distributive law with summands being products. (Contributed by AV, 25-Oct-2026)

Ref Expression
Hypotheses addmuldir1d.a ⊢ φ → A ∈ ℂ
addmuldir1d.b ⊢ φ → B ∈ ℂ
addmuldir1d.c ⊢ φ → C ∈ ℂ
addmuldir1d.d ⊢ φ → D ∈ ℂ
addmuldir1d.e ⊢ φ → E ∈ ℂ
addmuldir1d.f ⊢ φ → F ∈ ℂ
Assertion addmuldir1d ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ F = A ⁢ D + A ⁢ E ⁢ F + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ F

Proof

Step Hyp Ref Expression
1 addmuldir1d.a ⊢ φ → A ∈ ℂ
2 addmuldir1d.b ⊢ φ → B ∈ ℂ
3 addmuldir1d.c ⊢ φ → C ∈ ℂ
4 addmuldir1d.d ⊢ φ → D ∈ ℂ
5 addmuldir1d.e ⊢ φ → E ∈ ℂ
6 addmuldir1d.f ⊢ φ → F ∈ ℂ
7 2 3 mulcld ⊢ φ → B ⁢ C ∈ ℂ
8 5 6 mulcld ⊢ φ → E ⁢ F ∈ ℂ
9 1 7 4 8 muladdd ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ F = A ⁢ D + E ⁢ F ⁢ B ⁢ C + A ⁢ E ⁢ F + D ⁢ B ⁢ C
10 1 4 mulcld ⊢ φ → A ⁢ D ∈ ℂ
11 8 7 mulcld ⊢ φ → E ⁢ F ⁢ B ⁢ C ∈ ℂ
12 1 8 mulcld ⊢ φ → A ⁢ E ⁢ F ∈ ℂ
13 4 7 mulcld ⊢ φ → D ⁢ B ⁢ C ∈ ℂ
14 10 11 12 13 add42d ⊢ φ → A ⁢ D + E ⁢ F ⁢ B ⁢ C + A ⁢ E ⁢ F + D ⁢ B ⁢ C = A ⁢ D + A ⁢ E ⁢ F + D ⁢ B ⁢ C + E ⁢ F ⁢ B ⁢ C
15 8 7 mulcomd ⊢ φ → E ⁢ F ⁢ B ⁢ C = B ⁢ C ⁢ E ⁢ F
16 15 oveq2d ⊢ φ → D ⁢ B ⁢ C + E ⁢ F ⁢ B ⁢ C = D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ F
17 16 oveq2d ⊢ φ → A ⁢ D + A ⁢ E ⁢ F + D ⁢ B ⁢ C + E ⁢ F ⁢ B ⁢ C = A ⁢ D + A ⁢ E ⁢ F + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ F
18 9 14 17 3eqtrd ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ F = A ⁢ D + A ⁢ E ⁢ F + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ F