Metamath Proof Explorer


Theorem subscan2d

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 subscan2d ⊢ φ → A - s C = B - s C ↔ 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 1 3 subsvald ⊢ φ → A - s C = A + s + s ⁡ C
5 2 3 subsvald ⊢ φ → B - s C = B + s + s ⁡ C
6 4 5 eqeq12d ⊢ φ → A - s C = B - s C ↔ A + s + s ⁡ C = B + s + s ⁡ C
7 3 negscld ⊢ φ → + s ⁡ C ∈ No
8 1 2 7 addscan2d ⊢ φ → A + s + s ⁡ C = B + s + s ⁡ C ↔ A = B
9 6 8 bitrd ⊢ φ → A - s C = B - s C ↔ A = B