Theorem axextndbi 35456

Description: axextnd 10612 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 10612	. . 3 ⊢ ∃𝑧((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
2		elequ2 2113	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3	2	jctl 522	. . 3 ⊢ (((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
4	1, 3	eximii 1831	. 2 ⊢ ∃𝑧((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦))
5		dfbi2 473	. . 3 ⊢ ((𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
6	5	exbii 1842	. 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 230	1 ⊢ ∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 394  ∃wex 1773

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2166  ax-13 2365  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-nf 1778  df-clel 2802  df-nfc 2877

This theorem is referenced by: (None)

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