Metamath Proof Explorer


Theorem addmuldir2d

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

Ref Expression
Hypotheses addmuldir2d.a ⊢ φ → A ∈ ℂ
addmuldir2d.b ⊢ φ → B ∈ ℂ
addmuldir2d.c ⊢ φ → C ∈ ℂ
addmuldir2d.d ⊢ φ → D ∈ ℂ
addmuldir2d.e ⊢ φ → E ∈ ℂ
Assertion addmuldir2d ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ C = A ⁢ D + A ⁢ E + D ⁢ B ⁢ C + B ⁢ E ⁢ C 2

Proof

Step Hyp Ref Expression
1 addmuldir2d.a ⊢ φ → A ∈ ℂ
2 addmuldir2d.b ⊢ φ → B ∈ ℂ
3 addmuldir2d.c ⊢ φ → C ∈ ℂ
4 addmuldir2d.d ⊢ φ → D ∈ ℂ
5 addmuldir2d.e ⊢ φ → E ∈ ℂ
6 1 2 3 4 5 3 addmuldir1d ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ C = A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ C
7 1 4 mulcld ⊢ φ → A ⁢ D ∈ ℂ
8 5 3 mulcld ⊢ φ → E ⁢ C ∈ ℂ
9 1 8 mulcld ⊢ φ → A ⁢ E ⁢ C ∈ ℂ
10 7 9 addcld ⊢ φ → A ⁢ D + A ⁢ E ⁢ C ∈ ℂ
11 2 3 mulcld ⊢ φ → B ⁢ C ∈ ℂ
12 4 11 mulcld ⊢ φ → D ⁢ B ⁢ C ∈ ℂ
13 11 8 mulcld ⊢ φ → B ⁢ C ⁢ E ⁢ C ∈ ℂ
14 10 12 13 addassd ⊢ φ → A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ C = A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ C
15 7 9 12 addassd ⊢ φ → A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C = A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C
16 1 5 mulcld ⊢ φ → A ⁢ E ∈ ℂ
17 4 2 mulcld ⊢ φ → D ⁢ B ∈ ℂ
18 16 17 3 adddird ⊢ φ → A ⁢ E + D ⁢ B ⁢ C = A ⁢ E ⁢ C + D ⁢ B ⁢ C
19 1 5 3 mulassd ⊢ φ → A ⁢ E ⁢ C = A ⁢ E ⁢ C
20 4 2 3 mulassd ⊢ φ → D ⁢ B ⁢ C = D ⁢ B ⁢ C
21 19 20 oveq12d ⊢ φ → A ⁢ E ⁢ C + D ⁢ B ⁢ C = A ⁢ E ⁢ C + D ⁢ B ⁢ C
22 18 21 eqtr2d ⊢ φ → A ⁢ E ⁢ C + D ⁢ B ⁢ C = A ⁢ E + D ⁢ B ⁢ C
23 22 oveq2d ⊢ φ → A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C = A ⁢ D + A ⁢ E + D ⁢ B ⁢ C
24 15 23 eqtrd ⊢ φ → A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C = A ⁢ D + A ⁢ E + D ⁢ B ⁢ C
25 2 3 5 3 mul4d ⊢ φ → B ⁢ C ⁢ E ⁢ C = B ⁢ E ⁢ C ⁢ C
26 sqval ⊢ C ∈ ℂ → C 2 = C ⁢ C
27 26 eqcomd ⊢ C ∈ ℂ → C ⁢ C = C 2
28 3 27 syl ⊢ φ → C ⁢ C = C 2
29 28 oveq2d ⊢ φ → B ⁢ E ⁢ C ⁢ C = B ⁢ E ⁢ C 2
30 25 29 eqtrd ⊢ φ → B ⁢ C ⁢ E ⁢ C = B ⁢ E ⁢ C 2
31 24 30 oveq12d ⊢ φ → A ⁢ D + A ⁢ E ⁢ C + D ⁢ B ⁢ C + B ⁢ C ⁢ E ⁢ C = A ⁢ D + A ⁢ E + D ⁢ B ⁢ C + B ⁢ E ⁢ C 2
32 6 14 31 3eqtr2d ⊢ φ → A + B ⁢ C ⁢ D + E ⁢ C = A ⁢ D + A ⁢ E + D ⁢ B ⁢ C + B ⁢ E ⁢ C 2