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Theorem equsex 2450
Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2404. See equsexvw 2026 and equsexv 2304 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsal 2449. See equsexALT 2451 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 480 . . 3 ((𝑥 = 𝑦𝜑) → 𝜓)
41, 3exlimi 2253 . 2 (∃𝑥(𝑥 = 𝑦𝜑) → 𝜓)
51, 2equsal 2449 . . 3 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
6 equs4 2448 . . 3 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
75, 6sylbir 237 . 2 (𝜓 → ∃𝑥(𝑥 = 𝑦𝜑))
84, 7impbii 211 1 (∃𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 399  wal 1559  wex 1800  wnf 1804
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-12 2213  ax-13 2404
This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1801  df-nf 1805
This theorem is referenced by:  equsexh  2453  sb5rf  2499
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