Metamath Proof Explorer


Theorem oppcbas

Description: Base set of an opposite category. (Contributed by Mario Carneiro, 2-Jan-2017) (Proof shortened by AV, 18-Oct-2024)

Ref Expression
Hypotheses oppcbas.1 ⊢ O = oppCat ⁡ C
oppcbas.2 ⊢ B = Base C
Assertion oppcbas ⊢ B = Base O

Proof

Step Hyp Ref Expression
1 oppcbas.1 ⊢ O = oppCat ⁡ C
2 oppcbas.2 ⊢ B = Base C
3 baseid ⊢ Base = Slot Base ndx
4 slotsbhcdif ⊢ Base ndx ≠ Hom ⁡ ndx ∧ Base ndx ≠ comp ⁡ ndx ∧ Hom ⁡ ndx ≠ comp ⁡ ndx
5 4 simp1i ⊢ Base ndx ≠ Hom ⁡ ndx
6 3 5 setsnid ⊢ Base C = Base C sSet Hom ⁡ ndx tpos Hom ⁡ C
7 4 simp2i ⊢ Base ndx ≠ comp ⁡ ndx
8 3 7 setsnid ⊢ Base C sSet Hom ⁡ ndx tpos Hom ⁡ C = Base C sSet Hom ⁡ ndx tpos Hom ⁡ C sSet comp ⁡ ndx u ∈ Base C × Base C , z ∈ Base C ⟼ tpos z 2 nd ⁡ u comp ⁡ C 1 st ⁡ u
9 6 8 eqtri ⊢ Base C = Base C sSet Hom ⁡ ndx tpos Hom ⁡ C sSet comp ⁡ ndx u ∈ Base C × Base C , z ∈ Base C ⟼ tpos z 2 nd ⁡ u comp ⁡ C 1 st ⁡ u
10 eqid ⊢ Base C = Base C
11 eqid ⊢ Hom ⁡ C = Hom ⁡ C
12 eqid ⊢ comp ⁡ C = comp ⁡ C
13 10 11 12 1 oppcval ⊢ C ∈ V → O = C sSet Hom ⁡ ndx tpos Hom ⁡ C sSet comp ⁡ ndx u ∈ Base C × Base C , z ∈ Base C ⟼ tpos z 2 nd ⁡ u comp ⁡ C 1 st ⁡ u
14 13 fveq2d ⊢ C ∈ V → Base O = Base C sSet Hom ⁡ ndx tpos Hom ⁡ C sSet comp ⁡ ndx u ∈ Base C × Base C , z ∈ Base C ⟼ tpos z 2 nd ⁡ u comp ⁡ C 1 st ⁡ u
15 9 14 eqtr4id ⊢ C ∈ V → Base C = Base O
16 base0 ⊢ ∅ = Base ∅
17 16 eqcomi ⊢ Base ∅ = ∅
18 17 1 fveqprc ⊢ ¬ C ∈ V → Base C = Base O
19 15 18 pm2.61i ⊢ Base C = Base O
20 2 19 eqtri ⊢ B = Base O