Metamath Proof Explorer


Theorem efmndsgrp

Description: The monoid of endofunctions on a class A is a semigroup. (Contributed by AV, 28-Jan-2024)

Ref Expression
Hypothesis efmndmgm.g 𝐺 = ( EndoFMnd ‘ 𝐴 )
Assertion efmndsgrp 𝐺 ∈ Smgrp

Proof

Step Hyp Ref Expression
1 efmndmgm.g 𝐺 = ( EndoFMnd ‘ 𝐴 )
2 1 efmndmgm 𝐺 ∈ Mgm
3 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
4 eqid ( +g𝐺 ) = ( +g𝐺 )
5 1 3 4 efmndcl ( ( 𝑥 ∈ ( Base ‘ 𝐺 ) ∧ 𝑦 ∈ ( Base ‘ 𝐺 ) ) → ( 𝑥 ( +g𝐺 ) 𝑦 ) ∈ ( Base ‘ 𝐺 ) )
6 1 3 4 efmndov ( ( 𝑥 ∈ ( Base ‘ 𝐺 ) ∧ 𝑦 ∈ ( Base ‘ 𝐺 ) ) → ( 𝑥 ( +g𝐺 ) 𝑦 ) = ( 𝑥𝑦 ) )
7 5 6 symggrplem ( ( 𝑓 ∈ ( Base ‘ 𝐺 ) ∧ 𝑔 ∈ ( Base ‘ 𝐺 ) ∧ ∈ ( Base ‘ 𝐺 ) ) → ( ( 𝑓 ( +g𝐺 ) 𝑔 ) ( +g𝐺 ) ) = ( 𝑓 ( +g𝐺 ) ( 𝑔 ( +g𝐺 ) ) ) )
8 7 rgen3 𝑓 ∈ ( Base ‘ 𝐺 ) ∀ 𝑔 ∈ ( Base ‘ 𝐺 ) ∀ ∈ ( Base ‘ 𝐺 ) ( ( 𝑓 ( +g𝐺 ) 𝑔 ) ( +g𝐺 ) ) = ( 𝑓 ( +g𝐺 ) ( 𝑔 ( +g𝐺 ) ) )
9 3 4 issgrp ( 𝐺 ∈ Smgrp ↔ ( 𝐺 ∈ Mgm ∧ ∀ 𝑓 ∈ ( Base ‘ 𝐺 ) ∀ 𝑔 ∈ ( Base ‘ 𝐺 ) ∀ ∈ ( Base ‘ 𝐺 ) ( ( 𝑓 ( +g𝐺 ) 𝑔 ) ( +g𝐺 ) ) = ( 𝑓 ( +g𝐺 ) ( 𝑔 ( +g𝐺 ) ) ) ) )
10 2 8 9 mpbir2an 𝐺 ∈ Smgrp