Metamath Proof Explorer


Theorem subsge0d

Description: Non-negative subtraction. (Contributed by Scott Fenton, 26-May-2025)

Ref Expression
Hypotheses subsge0d.1
|- ( ph -> A e. No )
subsge0d.2
|- ( ph -> B e. No )
Assertion subsge0d
|- ( ph -> ( 0s <_s ( A -s B ) <-> B <_s A ) )

Proof

Step Hyp Ref Expression
1 subsge0d.1
 |-  ( ph -> A e. No )
2 subsge0d.2
 |-  ( ph -> B e. No )
3 0sno
 |-  0s e. No
4 3 a1i
 |-  ( ph -> 0s e. No )
5 1 2 subscld
 |-  ( ph -> ( A -s B ) e. No )
6 4 5 2 sleadd1d
 |-  ( ph -> ( 0s <_s ( A -s B ) <-> ( 0s +s B ) <_s ( ( A -s B ) +s B ) ) )
7 addslid
 |-  ( B e. No -> ( 0s +s B ) = B )
8 2 7 syl
 |-  ( ph -> ( 0s +s B ) = B )
9 npcans
 |-  ( ( A e. No /\ B e. No ) -> ( ( A -s B ) +s B ) = A )
10 1 2 9 syl2anc
 |-  ( ph -> ( ( A -s B ) +s B ) = A )
11 8 10 breq12d
 |-  ( ph -> ( ( 0s +s B ) <_s ( ( A -s B ) +s B ) <-> B <_s A ) )
12 6 11 bitrd
 |-  ( ph -> ( 0s <_s ( A -s B ) <-> B <_s A ) )