Metamath Proof Explorer


Theorem subscl

Description: Closure law for surreal subtraction. (Contributed by Scott Fenton, 3-Feb-2025)

Ref Expression
Assertion subscl ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) ∈ No )

Proof

Step Hyp Ref Expression
1 subsval ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
2 negscl ⊢ ( 𝐵 ∈ No → ( -us ‘ 𝐵 ) ∈ No )
3 addscl ⊢ ( ( 𝐴 ∈ No ∧ ( -us ‘ 𝐵 ) ∈ No ) → ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ No )
4 2 3 sylan2 ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ No )
5 1 4 eqeltrd ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ) → ( 𝐴 -s 𝐵 ) ∈ No )