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Theorem equsex 2416
Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2370. See equsexvw 2005 and equsexv 2269 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal 2415. See equsexALT 2417 for an alternate proof. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 6-Feb-2018.) (New usage is discouraged.)
Hypotheses
Ref Expression
equsal.1 𝑥𝜓
equsal.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
equsex (∃𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)

Proof of Theorem equsex
StepHypRef Expression
1 equsal.1 . . 3 𝑥𝜓
2 equsal.2 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
32biimpa 476 . . 3 ((𝑥 = 𝑦𝜑) → 𝜓)
41, 3exlimi 2218 . 2 (∃𝑥(𝑥 = 𝑦𝜑) → 𝜓)
51, 2equsal 2415 . . 3 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
6 equs4 2414 . . 3 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
75, 6sylbir 235 . 2 (𝜓 → ∃𝑥(𝑥 = 𝑦𝜑))
84, 7impbii 209 1 (∃𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wal 1538  wex 1779  wnf 1783
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-12 2178  ax-13 2370
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-nf 1784
This theorem is referenced by:  equsexh  2419  sb5rf  2465
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