Metamath Proof Explorer


Theorem djulf1o

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

Ref Expression
Assertion djulf1o ⊢ inl : V ⟶ 1-1 onto ∅ × V

Proof

Step Hyp Ref Expression
1 df-inl ⊢ inl = x ∈ V ⟼ ∅ x
2 0ex ⊢ ∅ ∈ V
3 2 snid ⊢ ∅ ∈ ∅
4 opelxpi ⊢ ∅ ∈ ∅ ∧ x ∈ V → ∅ x ∈ ∅ × V
5 3 4 mpan ⊢ x ∈ V → ∅ x ∈ ∅ × V
6 5 adantl ⊢ ⊤ ∧ x ∈ V → ∅ x ∈ ∅ × V
7 fvexd ⊢ ⊤ ∧ y ∈ ∅ × V → 2 nd ⁡ y ∈ V
8 1st2nd2 ⊢ y ∈ ∅ × V → y = 1 st ⁡ y 2 nd ⁡ y
9 xp1st ⊢ y ∈ ∅ × V → 1 st ⁡ y ∈ ∅
10 elsni ⊢ 1 st ⁡ y ∈ ∅ → 1 st ⁡ y = ∅
11 9 10 syl ⊢ y ∈ ∅ × V → 1 st ⁡ y = ∅
12 11 opeq1d ⊢ y ∈ ∅ × V → 1 st ⁡ y 2 nd ⁡ y = ∅ 2 nd ⁡ y
13 8 12 eqtrd ⊢ y ∈ ∅ × V → y = ∅ 2 nd ⁡ y
14 13 eqeq2d ⊢ y ∈ ∅ × V → ∅ x = y ↔ ∅ x = ∅ 2 nd ⁡ y
15 eqcom ⊢ ∅ x = y ↔ y = ∅ x
16 eqid ⊢ ∅ = ∅
17 vex ⊢ x ∈ V
18 2 17 opth ⊢ ∅ x = ∅ 2 nd ⁡ y ↔ ∅ = ∅ ∧ x = 2 nd ⁡ y
19 16 18 mpbiran ⊢ ∅ x = ∅ 2 nd ⁡ y ↔ x = 2 nd ⁡ y
20 14 15 19 3bitr3g ⊢ y ∈ ∅ × V → y = ∅ x ↔ x = 2 nd ⁡ y
21 20 bicomd ⊢ y ∈ ∅ × V → x = 2 nd ⁡ y ↔ y = ∅ x
22 21 ad2antll ⊢ ⊤ ∧ x ∈ V ∧ y ∈ ∅ × V → x = 2 nd ⁡ y ↔ y = ∅ x
23 1 6 7 22 f1o2d ⊢ ⊤ → inl : V ⟶ 1-1 onto ∅ × V
24 23 mptru ⊢ inl : V ⟶ 1-1 onto ∅ × V