Metamath Proof Explorer


Theorem negsval2

Description: Surreal negation in terms of subtraction. (Contributed by Scott Fenton, 15-Apr-2025)

Ref Expression
Assertion negsval2
|- ( A e. No -> ( -us ` A ) = ( 0s -s A ) )

Proof

Step Hyp Ref Expression
1 0sno
 |-  0s e. No
2 subsval
 |-  ( ( 0s e. No /\ A e. No ) -> ( 0s -s A ) = ( 0s +s ( -us ` A ) ) )
3 1 2 mpan
 |-  ( A e. No -> ( 0s -s A ) = ( 0s +s ( -us ` A ) ) )
4 negscl
 |-  ( A e. No -> ( -us ` A ) e. No )
5 addslid
 |-  ( ( -us ` A ) e. No -> ( 0s +s ( -us ` A ) ) = ( -us ` A ) )
6 4 5 syl
 |-  ( A e. No -> ( 0s +s ( -us ` A ) ) = ( -us ` A ) )
7 3 6 eqtr2d
 |-  ( A e. No -> ( -us ` A ) = ( 0s -s A ) )