Metamath Proof Explorer


Theorem fparlem2

Description: Lemma for fpar . (Contributed by NM, 22-Dec-2008) (Revised by Mario Carneiro, 28-Apr-2015)

Ref Expression
Assertion fparlem2 ⊢ 2 nd ↾ V × V -1 y = V × y

Proof

Step Hyp Ref Expression
1 fvres ⊢ x ∈ V × V → 2 nd ↾ V × V ⁡ x = 2 nd ⁡ x
2 1 eqeq1d ⊢ x ∈ V × V → 2 nd ↾ V × V ⁡ x = y ↔ 2 nd ⁡ x = y
3 vex ⊢ y ∈ V
4 3 elsn2 ⊢ 2 nd ⁡ x ∈ y ↔ 2 nd ⁡ x = y
5 fvex ⊢ 1 st ⁡ x ∈ V
6 5 biantrur ⊢ 2 nd ⁡ x ∈ y ↔ 1 st ⁡ x ∈ V ∧ 2 nd ⁡ x ∈ y
7 4 6 bitr3i ⊢ 2 nd ⁡ x = y ↔ 1 st ⁡ x ∈ V ∧ 2 nd ⁡ x ∈ y
8 2 7 bitrdi ⊢ x ∈ V × V → 2 nd ↾ V × V ⁡ x = y ↔ 1 st ⁡ x ∈ V ∧ 2 nd ⁡ x ∈ y
9 8 pm5.32i ⊢ x ∈ V × V ∧ 2 nd ↾ V × V ⁡ x = y ↔ x ∈ V × V ∧ 1 st ⁡ x ∈ V ∧ 2 nd ⁡ x ∈ y
10 f2ndres ⊢ 2 nd ↾ V × V : V × V ⟶ V
11 ffn ⊢ 2 nd ↾ V × V : V × V ⟶ V → 2 nd ↾ V × V Fn V × V
12 fniniseg ⊢ 2 nd ↾ V × V Fn V × V → x ∈ 2 nd ↾ V × V -1 y ↔ x ∈ V × V ∧ 2 nd ↾ V × V ⁡ x = y
13 10 11 12 mp2b ⊢ x ∈ 2 nd ↾ V × V -1 y ↔ x ∈ V × V ∧ 2 nd ↾ V × V ⁡ x = y
14 elxp7 ⊢ x ∈ V × y ↔ x ∈ V × V ∧ 1 st ⁡ x ∈ V ∧ 2 nd ⁡ x ∈ y
15 9 13 14 3bitr4i ⊢ x ∈ 2 nd ↾ V × V -1 y ↔ x ∈ V × y
16 15 eqriv ⊢ 2 nd ↾ V × V -1 y = V × y