Metamath Proof Explorer


Theorem abnex

Description: Sufficient condition for a class abstraction to be a proper class. Lemma for snnex and pwnex . See the comment of abnexg . (Contributed by BJ, 2-May-2021)

Ref Expression
Assertion abnex ( ∀ 𝑥 ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) → ¬ { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } ∈ V )

Proof

Step Hyp Ref Expression
1 vprc ⊢ ¬ V ∈ V
2 alral ⊢ ( ∀ 𝑥 ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) → ∀ 𝑥 ∈ V ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) )
3 rexv ⊢ ( ∃ 𝑥 ∈ V 𝑦 = 𝐹 ↔ ∃ 𝑥 𝑦 = 𝐹 )
4 3 bicomi ⊢ ( ∃ 𝑥 𝑦 = 𝐹 ↔ ∃ 𝑥 ∈ V 𝑦 = 𝐹 )
5 4 abbii ⊢ { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } = { 𝑦 ∣ ∃ 𝑥 ∈ V 𝑦 = 𝐹 }
6 5 eleq1i ⊢ ( { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } ∈ V ↔ { 𝑦 ∣ ∃ 𝑥 ∈ V 𝑦 = 𝐹 } ∈ V )
7 6 biimpi ⊢ ( { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } ∈ V → { 𝑦 ∣ ∃ 𝑥 ∈ V 𝑦 = 𝐹 } ∈ V )
8 abnexg ⊢ ( ∀ 𝑥 ∈ V ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) → ( { 𝑦 ∣ ∃ 𝑥 ∈ V 𝑦 = 𝐹 } ∈ V → V ∈ V ) )
9 2 7 8 syl2im ⊢ ( ∀ 𝑥 ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) → ( { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } ∈ V → V ∈ V ) )
10 1 9 mtoi ⊢ ( ∀ 𝑥 ( 𝐹 ∈ 𝑉 ∧ 𝑥 ∈ 𝐹 ) → ¬ { 𝑦 ∣ ∃ 𝑥 𝑦 = 𝐹 } ∈ V )