Metamath Proof Explorer


Theorem subscan1d

Description: Cancellation law for surreal subtraction. (Contributed by Scott Fenton, 7-Nov-2025)

Ref Expression
Hypotheses subscand.1 ⊢ φ → A ∈ No
subscand.2 ⊢ φ → B ∈ No
subscand.3 ⊢ φ → C ∈ No
Assertion subscan1d ⊢ φ → C - s A = C - s B ↔ A = B

Proof

Step Hyp Ref Expression
1 subscand.1 ⊢ φ → A ∈ No
2 subscand.2 ⊢ φ → B ∈ No
3 subscand.3 ⊢ φ → C ∈ No
4 3 1 subsvald ⊢ φ → C - s A = C + s + s ⁡ A
5 3 2 subsvald ⊢ φ → C - s B = C + s + s ⁡ B
6 4 5 eqeq12d ⊢ φ → C - s A = C - s B ↔ C + s + s ⁡ A = C + s + s ⁡ B
7 1 negscld ⊢ φ → + s ⁡ A ∈ No
8 2 negscld ⊢ φ → + s ⁡ B ∈ No
9 7 8 3 addscan1d ⊢ φ → C + s + s ⁡ A = C + s + s ⁡ B ↔ + s ⁡ A = + s ⁡ B
10 negs11 ⊢ A ∈ No ∧ B ∈ No → + s ⁡ A = + s ⁡ B ↔ A = B
11 1 2 10 syl2anc ⊢ φ → + s ⁡ A = + s ⁡ B ↔ A = B
12 6 9 11 3bitrd ⊢ φ → C - s A = C - s B ↔ A = B