Theorem axext4 2606

Description: A bidirectional version of Extensionality. Although this theorem "looks" like it is just a definition of equality, it requires the Axiom of Extensionality for its proof under our axiomatization. See the comments for ax-ext 2602 and df-cleq 2615. (Contributed by NM, 14-Nov-2008.)

Assertion

Ref	Expression
axext4	⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Proof of Theorem axext4

Step	Hyp	Ref	Expression
1		elequ2 2004	. . 3 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
2	1	alrimiv 1855	. 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3		axext3 2604	. 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
4	2, 3	impbii 199	1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   ↔ wb 196  ∀wal 1481

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705

This theorem is referenced by:  axc11next  38607