Metamath Proof Explorer


Theorem subscan1d

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

Ref Expression
Hypotheses subscand.1
|- ( ph -> A e. No )
subscand.2
|- ( ph -> B e. No )
subscand.3
|- ( ph -> C e. No )
Assertion subscan1d
|- ( ph -> ( ( C -s A ) = ( C -s B ) <-> A = B ) )

Proof

Step Hyp Ref Expression
1 subscand.1
 |-  ( ph -> A e. No )
2 subscand.2
 |-  ( ph -> B e. No )
3 subscand.3
 |-  ( ph -> C e. No )
4 3 1 subsvald
 |-  ( ph -> ( C -s A ) = ( C +s ( -us ` A ) ) )
5 3 2 subsvald
 |-  ( ph -> ( C -s B ) = ( C +s ( -us ` B ) ) )
6 4 5 eqeq12d
 |-  ( ph -> ( ( C -s A ) = ( C -s B ) <-> ( C +s ( -us ` A ) ) = ( C +s ( -us ` B ) ) ) )
7 1 negscld
 |-  ( ph -> ( -us ` A ) e. No )
8 2 negscld
 |-  ( ph -> ( -us ` B ) e. No )
9 7 8 3 addscan1d
 |-  ( ph -> ( ( C +s ( -us ` A ) ) = ( C +s ( -us ` B ) ) <-> ( -us ` A ) = ( -us ` B ) ) )
10 negs11
 |-  ( ( A e. No /\ B e. No ) -> ( ( -us ` A ) = ( -us ` B ) <-> A = B ) )
11 1 2 10 syl2anc
 |-  ( ph -> ( ( -us ` A ) = ( -us ` B ) <-> A = B ) )
12 6 9 11 3bitrd
 |-  ( ph -> ( ( C -s A ) = ( C -s B ) <-> A = B ) )