Theorem axextndbi 35805

Description: axextnd 10631 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 10631	. . 3 ⊢ ∃𝑧((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
2		elequ2 2123	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3	2	jctl 523	. . 3 ⊢ (((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
4	1, 3	eximii 1837	. 2 ⊢ ∃𝑧((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦))
5		dfbi2 474	. . 3 ⊢ ((𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
6	5	exbii 1848	. 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 231	1 ⊢ ∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∃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 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-13 2377  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-clel 2816  df-nfc 2892

This theorem is referenced by: (None)