Metamath Proof Explorer


Theorem efmnd0nmnd

Description: Even the monoid of endofunctions on the empty set is actually a monoid. (Contributed by AV, 31-Jan-2024)

Ref Expression
Assertion efmnd0nmnd ( EndoFMnd ‘ ∅ ) ∈ Mnd

Proof

Step Hyp Ref Expression
1 0ex ∅ ∈ V
2 eqid ( EndoFMnd ‘ ∅ ) = ( EndoFMnd ‘ ∅ )
3 2 efmndmnd ( ∅ ∈ V → ( EndoFMnd ‘ ∅ ) ∈ Mnd )
4 1 3 ax-mp ( EndoFMnd ‘ ∅ ) ∈ Mnd