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Theorem axextndbi 32659
 Description: axextnd 9859 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 9859 . . 3 𝑧((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)
2 elequ2 2096 . . . 4 (𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦))
32jctl 524 . . 3 (((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦) → ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
41, 3eximii 1818 . 2 𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦))
5 dfbi2 475 . . 3 ((𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
65exbii 1829 . 2 (∃𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦)) ↔ ∃𝑧((𝑥 = 𝑦 → (𝑧𝑥𝑧𝑦)) ∧ ((𝑧𝑥𝑧𝑦) → 𝑥 = 𝑦)))
74, 6mpbir 232 1 𝑧(𝑥 = 𝑦 ↔ (𝑧𝑥𝑧𝑦))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396  ∃wex 1761 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1777  ax-4 1791  ax-5 1888  ax-6 1947  ax-7 1992  ax-8 2083  ax-9 2091  ax-10 2112  ax-11 2126  ax-12 2141  ax-13 2344  ax-ext 2769 This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1525  df-ex 1762  df-nf 1766  df-nfc 2935 This theorem is referenced by: (None)
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