Metamath Proof Explorer


Theorem grpidcld

Description: The identity element of a group belongs to the group. (Contributed by Thierry Arnoux, 4-May-2026)

Ref Expression
Hypotheses grpidcld.1 𝐵 = ( Base ‘ 𝐺 )
grpidcld.2 0 = ( 0g𝐺 )
grpidcld.3 ( 𝜑𝐺 ∈ Grp )
Assertion grpidcld ( 𝜑0𝐵 )

Proof

Step Hyp Ref Expression
1 grpidcld.1 𝐵 = ( Base ‘ 𝐺 )
2 grpidcld.2 0 = ( 0g𝐺 )
3 grpidcld.3 ( 𝜑𝐺 ∈ Grp )
4 1 2 grpidcl ( 𝐺 ∈ Grp → 0𝐵 )
5 3 4 syl ( 𝜑0𝐵 )