Metamath Proof Explorer


Theorem negsubsdi2d

Description: Distribution of negative over subtraction. (Contributed by Scott Fenton, 5-Feb-2025)

Ref Expression
Hypotheses negsubsdi2d.1 ⊢ φ → A ∈ No
negsubsdi2d.2 ⊢ φ → B ∈ No
Assertion negsubsdi2d ⊢ φ → + s ⁡ A - s B = B - s A

Proof

Step Hyp Ref Expression
1 negsubsdi2d.1 ⊢ φ → A ∈ No
2 negsubsdi2d.2 ⊢ φ → B ∈ No
3 2 negscld ⊢ φ → + s ⁡ B ∈ No
4 negsdi ⊢ A ∈ No ∧ + s ⁡ B ∈ No → + s ⁡ A + s + s ⁡ B = + s ⁡ A + s + s ⁡ + s ⁡ B
5 1 3 4 syl2anc ⊢ φ → + s ⁡ A + s + s ⁡ B = + s ⁡ A + s + s ⁡ + s ⁡ B
6 negnegs ⊢ B ∈ No → + s ⁡ + s ⁡ B = B
7 2 6 syl ⊢ φ → + s ⁡ + s ⁡ B = B
8 7 oveq2d ⊢ φ → + s ⁡ A + s + s ⁡ + s ⁡ B = + s ⁡ A + s B
9 1 negscld ⊢ φ → + s ⁡ A ∈ No
10 9 2 addscomd ⊢ φ → + s ⁡ A + s B = B + s + s ⁡ A
11 5 8 10 3eqtrd ⊢ φ → + s ⁡ A + s + s ⁡ B = B + s + s ⁡ A
12 1 2 subsvald ⊢ φ → A - s B = A + s + s ⁡ B
13 12 fveq2d ⊢ φ → + s ⁡ A - s B = + s ⁡ A + s + s ⁡ B
14 2 1 subsvald ⊢ φ → B - s A = B + s + s ⁡ A
15 11 13 14 3eqtr4d ⊢ φ → + s ⁡ A - s B = B - s A