Metamath Proof Explorer


Theorem subscan1d

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

Ref Expression
Hypotheses subscand.1 ( 𝜑𝐴 No )
subscand.2 ( 𝜑𝐵 No )
subscand.3 ( 𝜑𝐶 No )
Assertion subscan1d ( 𝜑 → ( ( 𝐶 -s 𝐴 ) = ( 𝐶 -s 𝐵 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 subscand.1 ( 𝜑𝐴 No )
2 subscand.2 ( 𝜑𝐵 No )
3 subscand.3 ( 𝜑𝐶 No )
4 3 1 subsvald ( 𝜑 → ( 𝐶 -s 𝐴 ) = ( 𝐶 +s ( -us𝐴 ) ) )
5 3 2 subsvald ( 𝜑 → ( 𝐶 -s 𝐵 ) = ( 𝐶 +s ( -us𝐵 ) ) )
6 4 5 eqeq12d ( 𝜑 → ( ( 𝐶 -s 𝐴 ) = ( 𝐶 -s 𝐵 ) ↔ ( 𝐶 +s ( -us𝐴 ) ) = ( 𝐶 +s ( -us𝐵 ) ) ) )
7 1 negscld ( 𝜑 → ( -us𝐴 ) ∈ No )
8 2 negscld ( 𝜑 → ( -us𝐵 ) ∈ No )
9 7 8 3 addscan1d ( 𝜑 → ( ( 𝐶 +s ( -us𝐴 ) ) = ( 𝐶 +s ( -us𝐵 ) ) ↔ ( -us𝐴 ) = ( -us𝐵 ) ) )
10 negs11 ( ( 𝐴 No 𝐵 No ) → ( ( -us𝐴 ) = ( -us𝐵 ) ↔ 𝐴 = 𝐵 ) )
11 1 2 10 syl2anc ( 𝜑 → ( ( -us𝐴 ) = ( -us𝐵 ) ↔ 𝐴 = 𝐵 ) )
12 6 9 11 3bitrd ( 𝜑 → ( ( 𝐶 -s 𝐴 ) = ( 𝐶 -s 𝐵 ) ↔ 𝐴 = 𝐵 ) )