Metamath Proof Explorer


Theorem psrbagsn

Description: A singleton bag is a bag. (Contributed by Stefan O'Rear, 9-Mar-2015)

Ref Expression
Hypothesis psrbag0.d ⊢ D = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
Assertion psrbagsn ⊢ I ∈ V → x ∈ I ⟼ if x = K 1 0 ∈ D

Proof

Step Hyp Ref Expression
1 psrbag0.d ⊢ D = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
2 1nn0 ⊢ 1 ∈ ℕ 0
3 0nn0 ⊢ 0 ∈ ℕ 0
4 2 3 ifcli ⊢ if x = K 1 0 ∈ ℕ 0
5 4 a1i ⊢ ⊤ ∧ x ∈ I → if x = K 1 0 ∈ ℕ 0
6 5 fmpttd ⊢ ⊤ → x ∈ I ⟼ if x = K 1 0 : I ⟶ ℕ 0
7 6 mptru ⊢ x ∈ I ⟼ if x = K 1 0 : I ⟶ ℕ 0
8 eqid ⊢ x ∈ I ⟼ if x = K 1 0 = x ∈ I ⟼ if x = K 1 0
9 8 mptpreima ⊢ x ∈ I ⟼ if x = K 1 0 -1 ℕ = x ∈ I | if x = K 1 0 ∈ ℕ
10 snfi ⊢ K ∈ Fin
11 inss1 ⊢ x | x = K ∩ I ⊆ x | x = K
12 dfrab2 ⊢ x ∈ I | x = K = x | x = K ∩ I
13 df-sn ⊢ K = x | x = K
14 11 12 13 3sstr4i ⊢ x ∈ I | x = K ⊆ K
15 ssfi ⊢ K ∈ Fin ∧ x ∈ I | x = K ⊆ K → x ∈ I | x = K ∈ Fin
16 10 14 15 mp2an ⊢ x ∈ I | x = K ∈ Fin
17 0nnn ⊢ ¬ 0 ∈ ℕ
18 iffalse ⊢ ¬ x = K → if x = K 1 0 = 0
19 18 eleq1d ⊢ ¬ x = K → if x = K 1 0 ∈ ℕ ↔ 0 ∈ ℕ
20 17 19 mtbiri ⊢ ¬ x = K → ¬ if x = K 1 0 ∈ ℕ
21 20 con4i ⊢ if x = K 1 0 ∈ ℕ → x = K
22 21 a1i ⊢ x ∈ I → if x = K 1 0 ∈ ℕ → x = K
23 22 ss2rabi ⊢ x ∈ I | if x = K 1 0 ∈ ℕ ⊆ x ∈ I | x = K
24 ssfi ⊢ x ∈ I | x = K ∈ Fin ∧ x ∈ I | if x = K 1 0 ∈ ℕ ⊆ x ∈ I | x = K → x ∈ I | if x = K 1 0 ∈ ℕ ∈ Fin
25 16 23 24 mp2an ⊢ x ∈ I | if x = K 1 0 ∈ ℕ ∈ Fin
26 9 25 eqeltri ⊢ x ∈ I ⟼ if x = K 1 0 -1 ℕ ∈ Fin
27 7 26 pm3.2i ⊢ x ∈ I ⟼ if x = K 1 0 : I ⟶ ℕ 0 ∧ x ∈ I ⟼ if x = K 1 0 -1 ℕ ∈ Fin
28 1 psrbag ⊢ I ∈ V → x ∈ I ⟼ if x = K 1 0 ∈ D ↔ x ∈ I ⟼ if x = K 1 0 : I ⟶ ℕ 0 ∧ x ∈ I ⟼ if x = K 1 0 -1 ℕ ∈ Fin
29 27 28 mpbiri ⊢ I ∈ V → x ∈ I ⟼ if x = K 1 0 ∈ D