Metamath Proof Explorer


Theorem xpscf

Description: Equivalent condition for the pair function to be a proper function on A . (Contributed by Mario Carneiro, 20-Aug-2015)

Ref Expression
Assertion xpscf ⊢ ∅ X 1 𝑜 Y : 2 𝑜 ⟶ A ↔ X ∈ A ∧ Y ∈ A

Proof

Step Hyp Ref Expression
1 ifid ⊢ if k = ∅ A A = A
2 1 eleq2i ⊢ ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A ↔ ∅ X 1 𝑜 Y ⁡ k ∈ A
3 2 ralbii ⊢ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A ↔ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ A
4 3 anbi2i ⊢ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A ↔ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ A
5 df-3an ⊢ ∅ X 1 𝑜 Y ∈ V ∧ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A ↔ ∅ X 1 𝑜 Y ∈ V ∧ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A
6 elixp2 ⊢ ∅ X 1 𝑜 Y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A A ↔ ∅ X 1 𝑜 Y ∈ V ∧ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A
7 2onn ⊢ 2 𝑜 ∈ ω
8 fnex ⊢ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ 2 𝑜 ∈ ω → ∅ X 1 𝑜 Y ∈ V
9 7 8 mpan2 ⊢ ∅ X 1 𝑜 Y Fn 2 𝑜 → ∅ X 1 𝑜 Y ∈ V
10 9 pm4.71ri ⊢ ∅ X 1 𝑜 Y Fn 2 𝑜 ↔ ∅ X 1 𝑜 Y ∈ V ∧ ∅ X 1 𝑜 Y Fn 2 𝑜
11 10 anbi1i ⊢ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A ↔ ∅ X 1 𝑜 Y ∈ V ∧ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A
12 5 6 11 3bitr4i ⊢ ∅ X 1 𝑜 Y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A A ↔ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ if k = ∅ A A
13 ffnfv ⊢ ∅ X 1 𝑜 Y : 2 𝑜 ⟶ A ↔ ∅ X 1 𝑜 Y Fn 2 𝑜 ∧ ∀ k ∈ 2 𝑜 ∅ X 1 𝑜 Y ⁡ k ∈ A
14 4 12 13 3bitr4i ⊢ ∅ X 1 𝑜 Y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A A ↔ ∅ X 1 𝑜 Y : 2 𝑜 ⟶ A
15 xpsfrnel2 ⊢ ∅ X 1 𝑜 Y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A A ↔ X ∈ A ∧ Y ∈ A
16 14 15 bitr3i ⊢ ∅ X 1 𝑜 Y : 2 𝑜 ⟶ A ↔ X ∈ A ∧ Y ∈ A