Theorem axextb 2702

Description: A bidirectional version of the axiom of extensionality. Although this theorem looks like a definition of equality, it requires the axiom of extensionality for its proof under our axiomatization. See the comments for ax-ext 2699 and df-cleq 2720. (Contributed by NM, 14-Nov-2008.)

Assertion

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

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Proof of Theorem axextb

Step	Hyp	Ref	Expression
1		elequ2g 2115	. 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
2		axextg 2701	. 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
3	1, 2	impbii 208	1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   ↔ wb 205  ∀wal 1532

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1775

This theorem is referenced by:  dfcleq  2721  axc11next  43843