Metamath Proof Explorer


Theorem subscl

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

Ref Expression
Assertion subscl
|- ( ( A e. No /\ B e. No ) -> ( A -s B ) e. No )

Proof

Step Hyp Ref Expression
1 subsval
 |-  ( ( A e. No /\ B e. No ) -> ( A -s B ) = ( A +s ( -us ` B ) ) )
2 negscl
 |-  ( B e. No -> ( -us ` B ) e. No )
3 addscl
 |-  ( ( A e. No /\ ( -us ` B ) e. No ) -> ( A +s ( -us ` B ) ) e. No )
4 2 3 sylan2
 |-  ( ( A e. No /\ B e. No ) -> ( A +s ( -us ` B ) ) e. No )
5 1 4 eqeltrd
 |-  ( ( A e. No /\ B e. No ) -> ( A -s B ) e. No )