Metamath Proof Explorer


Theorem axacndlem3

Description: Lemma for the Axiom of Choice with no distinct variable conditions. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 3-Jan-2002) (New usage is discouraged.)

Ref Expression
Assertion axacndlem3 ( ∀ 𝑦 𝑦 = 𝑧 → ∃ 𝑥 ∀ 𝑦 ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )

Proof

Step Hyp Ref Expression
1 nfae ⊢ Ⅎ 𝑧 ∀ 𝑦 𝑦 = 𝑧
2 simpl ⊢ ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → 𝑦 ∈ 𝑧 )
3 2 alimi ⊢ ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∀ 𝑥 𝑦 ∈ 𝑧 )
4 nd3 ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ¬ ∀ 𝑥 𝑦 ∈ 𝑧 )
5 4 pm2.21d ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 𝑦 ∈ 𝑧 → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
6 3 5 syl5 ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
7 1 6 alrimi ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
8 7 axc4i ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ∀ 𝑦 ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
9 8 19.8ad ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ∃ 𝑥 ∀ 𝑦 ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )