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 ( -us ‘ 𝐴 ) ) )
5 3 2 subsvald ⊢ ( 𝜑 → ( 𝐶 -s 𝐵 ) = ( 𝐶 +s ( -us ‘ 𝐵 ) ) )
6 4 5 eqeq12d ⊢ ( 𝜑 → ( ( 𝐶 -s 𝐴 ) = ( 𝐶 -s 𝐵 ) ↔ ( 𝐶 +s ( -us ‘ 𝐴 ) ) = ( 𝐶 +s ( -us ‘ 𝐵 ) ) ) )
7 1 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐴 ) ∈ No )
8 2 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐵 ) ∈ No )
9 7 8 3 addscan1d ⊢ ( 𝜑 → ( ( 𝐶 +s ( -us ‘ 𝐴 ) ) = ( 𝐶 +s ( -us ‘ 𝐵 ) ) ↔ ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) ) )
10 negs11 ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) ↔ 𝐴 = 𝐵 ) )
11 1 2 10 syl2anc ⊢ ( 𝜑 → ( ( -us ‘ 𝐴 ) = ( -us ‘ 𝐵 ) ↔ 𝐴 = 𝐵 ) )
12 6 9 11 3bitrd ⊢ ( 𝜑 → ( ( 𝐶 -s 𝐴 ) = ( 𝐶 -s 𝐵 ) ↔ 𝐴 = 𝐵 ) )