Metamath Proof Explorer


Theorem brssc

Description: The subcategory subset relation is a relation. (Contributed by Mario Carneiro, 6-Jan-2017)

Ref Expression
Assertion brssc ⊢ H ⊆ cat J ↔ ∃ t J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x

Proof

Step Hyp Ref Expression
1 sscrel ⊢ Rel ⁡ ⊆ cat
2 1 brrelex12i ⊢ H ⊆ cat J → H ∈ V ∧ J ∈ V
3 vex ⊢ t ∈ V
4 3 3 xpex ⊢ t × t ∈ V
5 fnex ⊢ J Fn t × t ∧ t × t ∈ V → J ∈ V
6 4 5 mpan2 ⊢ J Fn t × t → J ∈ V
7 elex ⊢ H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x → H ∈ V
8 7 rexlimivw ⊢ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x → H ∈ V
9 6 8 anim12ci ⊢ J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x → H ∈ V ∧ J ∈ V
10 9 exlimiv ⊢ ∃ t J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x → H ∈ V ∧ J ∈ V
11 simpr ⊢ h = H ∧ j = J → j = J
12 11 fneq1d ⊢ h = H ∧ j = J → j Fn t × t ↔ J Fn t × t
13 simpl ⊢ h = H ∧ j = J → h = H
14 11 fveq1d ⊢ h = H ∧ j = J → j ⁡ x = J ⁡ x
15 14 pweqd ⊢ h = H ∧ j = J → 𝒫 j ⁡ x = 𝒫 J ⁡ x
16 15 ixpeq2dv ⊢ h = H ∧ j = J → ⨉ x ∈ s × s 𝒫 j ⁡ x = ⨉ x ∈ s × s 𝒫 J ⁡ x
17 13 16 eleq12d ⊢ h = H ∧ j = J → h ∈ ⨉ x ∈ s × s 𝒫 j ⁡ x ↔ H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x
18 17 rexbidv ⊢ h = H ∧ j = J → ∃ s ∈ 𝒫 t h ∈ ⨉ x ∈ s × s 𝒫 j ⁡ x ↔ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x
19 12 18 anbi12d ⊢ h = H ∧ j = J → j Fn t × t ∧ ∃ s ∈ 𝒫 t h ∈ ⨉ x ∈ s × s 𝒫 j ⁡ x ↔ J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x
20 19 exbidv ⊢ h = H ∧ j = J → ∃ t j Fn t × t ∧ ∃ s ∈ 𝒫 t h ∈ ⨉ x ∈ s × s 𝒫 j ⁡ x ↔ ∃ t J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x
21 df-ssc ⊢ ⊆ cat = h j | ∃ t j Fn t × t ∧ ∃ s ∈ 𝒫 t h ∈ ⨉ x ∈ s × s 𝒫 j ⁡ x
22 20 21 brabga ⊢ H ∈ V ∧ J ∈ V → H ⊆ cat J ↔ ∃ t J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x
23 2 10 22 pm5.21nii ⊢ H ⊆ cat J ↔ ∃ t J Fn t × t ∧ ∃ s ∈ 𝒫 t H ∈ ⨉ x ∈ s × s 𝒫 J ⁡ x