Metamath Proof Explorer


Theorem snsymgefmndeq

Description: The symmetric group on a singleton A is identical with the monoid of endofunctions on A . (Contributed by AV, 31-Mar-2024)

Ref Expression
Assertion snsymgefmndeq ⊢ A = X → EndoFMnd ⁡ A = SymGrp ⁡ A

Proof

Step Hyp Ref Expression
1 ssidd ⊢ X ∈ V → X X ⊆ X X
2 eqid ⊢ EndoFMnd ⁡ X = EndoFMnd ⁡ X
3 eqid ⊢ Base EndoFMnd ⁡ X = Base EndoFMnd ⁡ X
4 eqid ⊢ X = X
5 2 3 4 efmnd1bas ⊢ X ∈ V → Base EndoFMnd ⁡ X = X X
6 eqid ⊢ SymGrp ⁡ X = SymGrp ⁡ X
7 eqid ⊢ Base SymGrp ⁡ X = Base SymGrp ⁡ X
8 6 7 4 symg1bas ⊢ X ∈ V → Base SymGrp ⁡ X = X X
9 1 5 8 3sstr4d ⊢ X ∈ V → Base EndoFMnd ⁡ X ⊆ Base SymGrp ⁡ X
10 fvexd ⊢ X ∈ V → EndoFMnd ⁡ X ∈ V
11 fvexd ⊢ X ∈ V → Base SymGrp ⁡ X ∈ V
12 6 7 2 symgressbas ⊢ SymGrp ⁡ X = EndoFMnd ⁡ X ↾ 𝑠 Base SymGrp ⁡ X
13 12 3 ressid2 ⊢ Base EndoFMnd ⁡ X ⊆ Base SymGrp ⁡ X ∧ EndoFMnd ⁡ X ∈ V ∧ Base SymGrp ⁡ X ∈ V → SymGrp ⁡ X = EndoFMnd ⁡ X
14 9 10 11 13 syl3anc ⊢ X ∈ V → SymGrp ⁡ X = EndoFMnd ⁡ X
15 14 eqcomd ⊢ X ∈ V → EndoFMnd ⁡ X = SymGrp ⁡ X
16 fveq2 ⊢ A = X → EndoFMnd ⁡ A = EndoFMnd ⁡ X
17 fveq2 ⊢ A = X → SymGrp ⁡ A = SymGrp ⁡ X
18 16 17 eqeq12d ⊢ A = X → EndoFMnd ⁡ A = SymGrp ⁡ A ↔ EndoFMnd ⁡ X = SymGrp ⁡ X
19 15 18 syl5ibrcom ⊢ X ∈ V → A = X → EndoFMnd ⁡ A = SymGrp ⁡ A
20 snprc ⊢ ¬ X ∈ V ↔ X = ∅
21 20 biimpi ⊢ ¬ X ∈ V → X = ∅
22 21 eqeq2d ⊢ ¬ X ∈ V → A = X ↔ A = ∅
23 0symgefmndeq ⊢ EndoFMnd ⁡ ∅ = SymGrp ⁡ ∅
24 fveq2 ⊢ A = ∅ → EndoFMnd ⁡ A = EndoFMnd ⁡ ∅
25 fveq2 ⊢ A = ∅ → SymGrp ⁡ A = SymGrp ⁡ ∅
26 23 24 25 3eqtr4a ⊢ A = ∅ → EndoFMnd ⁡ A = SymGrp ⁡ A
27 22 26 biimtrdi ⊢ ¬ X ∈ V → A = X → EndoFMnd ⁡ A = SymGrp ⁡ A
28 19 27 pm2.61i ⊢ A = X → EndoFMnd ⁡ A = SymGrp ⁡ A