Metamath Proof Explorer


Theorem axacndlem2

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 axacndlem2 ( ∀ 𝑥 𝑥 = 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 𝑦 ∀ 𝑥 𝑥 = 𝑧
2 nfae ⊢ Ⅎ 𝑧 ∀ 𝑥 𝑥 = 𝑧
3 simpr ⊢ ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → 𝑧 ∈ 𝑤 )
4 3 alimi ⊢ ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∀ 𝑥 𝑧 ∈ 𝑤 )
5 nd1 ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ¬ ∀ 𝑥 𝑧 ∈ 𝑤 )
6 5 pm2.21d ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ( ∀ 𝑥 𝑧 ∈ 𝑤 → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
7 4 6 syl5 ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
8 2 7 alrimi ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
9 1 8 alrimi ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ∀ 𝑦 ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )
10 9 19.8ad ⊢ ( ∀ 𝑥 𝑥 = 𝑧 → ∃ 𝑥 ∀ 𝑦 ∀ 𝑧 ( ∀ 𝑥 ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) → ∃ 𝑤 ∀ 𝑦 ( ∃ 𝑤 ( ( 𝑦 ∈ 𝑧 ∧ 𝑧 ∈ 𝑤 ) ∧ ( 𝑦 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) ) ↔ 𝑦 = 𝑤 ) ) )