Metamath Proof Explorer


Theorem grpmnd

Description: A group is a monoid. (Contributed by Mario Carneiro, 6-Jan-2015)

Ref Expression
Assertion grpmnd ( 𝐺 ∈ Grp → 𝐺 ∈ Mnd )

Proof

Step Hyp Ref Expression
1 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
2 eqid ( +g𝐺 ) = ( +g𝐺 )
3 eqid ( 0g𝐺 ) = ( 0g𝐺 )
4 1 2 3 isgrp ( 𝐺 ∈ Grp ↔ ( 𝐺 ∈ Mnd ∧ ∀ 𝑎 ∈ ( Base ‘ 𝐺 ) ∃ 𝑚 ∈ ( Base ‘ 𝐺 ) ( 𝑚 ( +g𝐺 ) 𝑎 ) = ( 0g𝐺 ) ) )
5 4 simplbi ( 𝐺 ∈ Grp → 𝐺 ∈ Mnd )