Metamath Proof Explorer


Theorem affineequiv4

Description: Equivalence between two ways of expressing A as an affine combination of B and C . (Contributed by AV, 22-Jan-2023)

Ref Expression
Hypotheses affineequiv.a ⊢ φ → A ∈ ℂ
affineequiv.b ⊢ φ → B ∈ ℂ
affineequiv.c ⊢ φ → C ∈ ℂ
affineequiv.d ⊢ φ → D ∈ ℂ
Assertion affineequiv4 ⊢ φ → A = 1 − D ⁢ B + D ⁢ C ↔ A = D ⁢ C − B + B

Proof

Step Hyp Ref Expression
1 affineequiv.a ⊢ φ → A ∈ ℂ
2 affineequiv.b ⊢ φ → B ∈ ℂ
3 affineequiv.c ⊢ φ → C ∈ ℂ
4 affineequiv.d ⊢ φ → D ∈ ℂ
5 1 2 3 4 affineequiv3 ⊢ φ → A = 1 − D ⁢ B + D ⁢ C ↔ A − B = D ⁢ C − B
6 3 2 subcld ⊢ φ → C − B ∈ ℂ
7 4 6 mulcld ⊢ φ → D ⁢ C − B ∈ ℂ
8 1 2 7 subadd2d ⊢ φ → A − B = D ⁢ C − B ↔ D ⁢ C − B + B = A
9 eqcom ⊢ D ⁢ C − B + B = A ↔ A = D ⁢ C − B + B
10 8 9 bitrdi ⊢ φ → A − B = D ⁢ C − B ↔ A = D ⁢ C − B + B
11 5 10 bitrd ⊢ φ → A = 1 − D ⁢ B + D ⁢ C ↔ A = D ⁢ C − B + B