Theorem axextndbi 33053

Description: axextnd 10016 as a biconditional. (Contributed by Scott Fenton, 14-Dec-2010.)

Assertion

Ref	Expression
axextndbi	⊢ ∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Proof of Theorem axextndbi

Step	Hyp	Ref	Expression
1		axextnd 10016	. . 3 ⊢ ∃𝑧((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
2		elequ2 2128	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3	2	jctl 526	. . 3 ⊢ (((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
4	1, 3	eximii 1836	. 2 ⊢ ∃𝑧((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦))
5		dfbi2 477	. . 3 ⊢ ((𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
6	5	exbii 1847	. 2 ⊢ (∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ↔ ∃𝑧((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
7	4, 6	mpbir 233	1 ⊢ ∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∃wex 1779

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-13 2389  ax-ext 2796

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1539  df-ex 1780  df-nf 1784  df-nfc 2966

This theorem is referenced by: (None)