Theorem axext4 2807

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 2803 and df-cleq 2818. (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 2180	. . 3 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
2	1	alrimiv 2028	. 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3		axext3 2805	. 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
4	2, 3	impbii 201	1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   ↔ wb 198  ∀wal 1656

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-ext 2803

This theorem depends on definitions:  df-bi 199  df-an 387  df-ex 1881

This theorem is referenced by:  axc11next  39446