Metamath Proof Explorer


Theorem subsubs4d

Description: Law for double surreal subtraction. (Contributed by Scott Fenton, 9-Mar-2025)

Ref Expression
Hypotheses subsubs4d.1 ⊢ φ → A ∈ No
subsubs4d.2 ⊢ φ → B ∈ No
subsubs4d.3 ⊢ φ → C ∈ No
Assertion subsubs4d ⊢ φ → A - s B - s C = A - s B + s C

Proof

Step Hyp Ref Expression
1 subsubs4d.1 ⊢ φ → A ∈ No
2 subsubs4d.2 ⊢ φ → B ∈ No
3 subsubs4d.3 ⊢ φ → C ∈ No
4 2 negscld ⊢ φ → + s ⁡ B ∈ No
5 3 negscld ⊢ φ → + s ⁡ C ∈ No
6 1 4 5 addsassd ⊢ φ → A + s + s ⁡ B + s + s ⁡ C = A + s + s ⁡ B + s + s ⁡ C
7 1 2 subsvald ⊢ φ → A - s B = A + s + s ⁡ B
8 7 oveq1d ⊢ φ → A - s B - s C = A + s + s ⁡ B - s C
9 1 4 addscld ⊢ φ → A + s + s ⁡ B ∈ No
10 9 3 subsvald ⊢ φ → A + s + s ⁡ B - s C = A + s + s ⁡ B + s + s ⁡ C
11 8 10 eqtrd ⊢ φ → A - s B - s C = A + s + s ⁡ B + s + s ⁡ C
12 2 3 addscld ⊢ φ → B + s C ∈ No
13 1 12 subsvald ⊢ φ → A - s B + s C = A + s + s ⁡ B + s C
14 negsdi ⊢ B ∈ No ∧ C ∈ No → + s ⁡ B + s C = + s ⁡ B + s + s ⁡ C
15 2 3 14 syl2anc ⊢ φ → + s ⁡ B + s C = + s ⁡ B + s + s ⁡ C
16 15 oveq2d ⊢ φ → A + s + s ⁡ B + s C = A + s + s ⁡ B + s + s ⁡ C
17 13 16 eqtrd ⊢ φ → A - s B + s C = A + s + s ⁡ B + s + s ⁡ C
18 6 11 17 3eqtr4d ⊢ φ → A - s B - s C = A - s B + s C