Metamath Proof Explorer


Theorem djurf1o

Description: The right injection function on all sets is one to one and onto. (Contributed by Jim Kingdon, 22-Jun-2022)

Ref Expression
Assertion djurf1o ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V

Proof

Step Hyp Ref Expression
1 df-inr ⊢ inr = x ∈ V ⟼ 1 𝑜 x
2 1onn ⊢ 1 𝑜 ∈ ω
3 snidg ⊢ 1 𝑜 ∈ ω → 1 𝑜 ∈ 1 𝑜
4 2 3 ax-mp ⊢ 1 𝑜 ∈ 1 𝑜
5 opelxpi ⊢ 1 𝑜 ∈ 1 𝑜 ∧ x ∈ V → 1 𝑜 x ∈ 1 𝑜 × V
6 4 5 mpan ⊢ x ∈ V → 1 𝑜 x ∈ 1 𝑜 × V
7 6 adantl ⊢ ⊤ ∧ x ∈ V → 1 𝑜 x ∈ 1 𝑜 × V
8 fvexd ⊢ ⊤ ∧ y ∈ 1 𝑜 × V → 2 nd ⁡ y ∈ V
9 1st2nd2 ⊢ y ∈ 1 𝑜 × V → y = 1 st ⁡ y 2 nd ⁡ y
10 xp1st ⊢ y ∈ 1 𝑜 × V → 1 st ⁡ y ∈ 1 𝑜
11 elsni ⊢ 1 st ⁡ y ∈ 1 𝑜 → 1 st ⁡ y = 1 𝑜
12 10 11 syl ⊢ y ∈ 1 𝑜 × V → 1 st ⁡ y = 1 𝑜
13 12 opeq1d ⊢ y ∈ 1 𝑜 × V → 1 st ⁡ y 2 nd ⁡ y = 1 𝑜 2 nd ⁡ y
14 9 13 eqtrd ⊢ y ∈ 1 𝑜 × V → y = 1 𝑜 2 nd ⁡ y
15 14 eqeq2d ⊢ y ∈ 1 𝑜 × V → 1 𝑜 x = y ↔ 1 𝑜 x = 1 𝑜 2 nd ⁡ y
16 eqcom ⊢ 1 𝑜 x = y ↔ y = 1 𝑜 x
17 eqid ⊢ 1 𝑜 = 1 𝑜
18 1oex ⊢ 1 𝑜 ∈ V
19 vex ⊢ x ∈ V
20 18 19 opth ⊢ 1 𝑜 x = 1 𝑜 2 nd ⁡ y ↔ 1 𝑜 = 1 𝑜 ∧ x = 2 nd ⁡ y
21 17 20 mpbiran ⊢ 1 𝑜 x = 1 𝑜 2 nd ⁡ y ↔ x = 2 nd ⁡ y
22 15 16 21 3bitr3g ⊢ y ∈ 1 𝑜 × V → y = 1 𝑜 x ↔ x = 2 nd ⁡ y
23 22 bicomd ⊢ y ∈ 1 𝑜 × V → x = 2 nd ⁡ y ↔ y = 1 𝑜 x
24 23 ad2antll ⊢ ⊤ ∧ x ∈ V ∧ y ∈ 1 𝑜 × V → x = 2 nd ⁡ y ↔ y = 1 𝑜 x
25 1 7 8 24 f1o2d ⊢ ⊤ → inr : V ⟶ 1-1 onto 1 𝑜 × V
26 25 mptru ⊢ inr : V ⟶ 1-1 onto 1 𝑜 × V