Theorem axextndbi 36389

Description: axextnd 10604 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 10604	. . 3 ⊢ ∃𝑧((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
2		elequ2 2160	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
3	2	jctl 533	. . 3 ⊢ (((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
4	1, 3	eximii 1870	. 2 ⊢ ∃𝑧((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦))
5		dfbi2 480	. . 3 ⊢ ((𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) ∧ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)))
6	5	exbii 1881	. 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 234	1 ⊢ ∃𝑧(𝑥 = 𝑦 ↔ (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-13 2403  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-clel 2837  df-nfc 2911

This theorem is used by: (None)