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Theorem axextndbi 31684
Description: axextnd 9398 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 9398 . . 3 𝑧((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)
2 elequ2 2002 . . . 4 (𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦))
32jctl 563 . . 3 (((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
41, 3eximii 1762 . 2 𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦))
5 dfbi2 659 . . 3 ((𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
65exbii 1772 . 2 (∃𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ∃𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
74, 6mpbir 221 1 𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 384  wex 1702
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-8 1990  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1484  df-ex 1703  df-nf 1708  df-cleq 2613  df-clel 2616  df-nfc 2751
This theorem is referenced by: (None)
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