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