Metamath Proof Explorer


Theorem snhesn

Description: Any singleton is hereditary in any singleton. (Contributed by RP, 28-Mar-2020)

Ref Expression
Assertion snhesn ⊢ A A hereditary B

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 elima3 ⊢ x ∈ A A B ↔ ∃ y y ∈ B ∧ y x ∈ A A
3 velsn ⊢ x ∈ B ↔ x = B
4 2 3 imbi12i ⊢ x ∈ A A B → x ∈ B ↔ ∃ y y ∈ B ∧ y x ∈ A A → x = B
5 4 albii ⊢ ∀ x x ∈ A A B → x ∈ B ↔ ∀ x ∃ y y ∈ B ∧ y x ∈ A A → x = B
6 velsn ⊢ y ∈ B ↔ y = B
7 opex ⊢ y x ∈ V
8 7 elsn ⊢ y x ∈ A A ↔ y x = A A
9 vex ⊢ y ∈ V
10 9 1 opth ⊢ y x = A A ↔ y = A ∧ x = A
11 8 10 bitri ⊢ y x ∈ A A ↔ y = A ∧ x = A
12 6 11 anbi12i ⊢ y ∈ B ∧ y x ∈ A A ↔ y = B ∧ y = A ∧ x = A
13 3anass ⊢ y = B ∧ y = A ∧ x = A ↔ y = B ∧ y = A ∧ x = A
14 12 13 bitr4i ⊢ y ∈ B ∧ y x ∈ A A ↔ y = B ∧ y = A ∧ x = A
15 simp3 ⊢ y = B ∧ y = A ∧ x = A → x = A
16 simp2 ⊢ y = B ∧ y = A ∧ x = A → y = A
17 simp1 ⊢ y = B ∧ y = A ∧ x = A → y = B
18 15 16 17 3eqtr2d ⊢ y = B ∧ y = A ∧ x = A → x = B
19 14 18 sylbi ⊢ y ∈ B ∧ y x ∈ A A → x = B
20 19 exlimiv ⊢ ∃ y y ∈ B ∧ y x ∈ A A → x = B
21 5 20 mpgbir ⊢ ∀ x x ∈ A A B → x ∈ B
22 df-ss ⊢ A A B ⊆ B ↔ ∀ x x ∈ A A B → x ∈ B
23 21 22 mpbir ⊢ A A B ⊆ B
24 df-he ⊢ A A hereditary B ↔ A A B ⊆ B
25 23 24 mpbir ⊢ A A hereditary B