Metamath Proof Explorer


Theorem subgsubcld

Description: A subgroup is closed under group subtraction. (Contributed by Thierry Arnoux, 6-Jul-2025)

Ref Expression
Hypotheses subgsubcld.m
|- .- = ( -g ` G )
subgsubcld.s
|- ( ph -> S e. ( SubGrp ` G ) )
subgsubcld.x
|- ( ph -> X e. S )
subgsubcld.y
|- ( ph -> Y e. S )
Assertion subgsubcld
|- ( ph -> ( X .- Y ) e. S )

Proof

Step Hyp Ref Expression
1 subgsubcld.m
 |-  .- = ( -g ` G )
2 subgsubcld.s
 |-  ( ph -> S e. ( SubGrp ` G ) )
3 subgsubcld.x
 |-  ( ph -> X e. S )
4 subgsubcld.y
 |-  ( ph -> Y e. S )
5 1 subgsubcl
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S /\ Y e. S ) -> ( X .- Y ) e. S )
6 2 3 4 5 syl3anc
 |-  ( ph -> ( X .- Y ) e. S )