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Theorem axextndbi 33357
Description: axextnd 10094 as a biconditional. (Contributed by Scott Fenton, 14-Dec-2010.)
Assertion
Ref Expression
axextndbi 𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦))

Proof of Theorem axextndbi
StepHypRef Expression
1 axextnd 10094 . . 3 𝑧((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)
2 elequ2 2129 . . . 4 (𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦))
32jctl 527 . . 3 (((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
41, 3eximii 1843 . 2 𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦))
5 dfbi2 478 . . 3 ((𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
65exbii 1854 . 2 (∃𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ∃𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
74, 6mpbir 234 1 𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 399  wex 1786
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2162  ax-12 2179  ax-13 2373  ax-ext 2711
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-ex 1787  df-nf 1791  df-clel 2812  df-nfc 2882
This theorem is referenced by: (None)
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