Users' Mathboxes Mathbox for Scott Fenton < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  axextndbi Structured version   Visualization version   GIF version

Theorem axextndbi 30757
Description: axextnd 9266 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 9266 . . 3 𝑧((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)
2 elequ2 1990 . . . 4 (𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦))
32jctl 561 . . 3 (((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
41, 3eximii 1753 . 2 𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦))
5 dfbi2 657 . . 3 ((𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
65exbii 1763 . 2 (∃𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ∃𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
74, 6mpbir 219 1 𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 194  wa 382  wex 1694
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-8 1978  ax-9 1985  ax-10 2005  ax-11 2020  ax-12 2032  ax-13 2229  ax-ext 2586
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-cleq 2599  df-clel 2602  df-nfc 2736
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator