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Theorem equsal 2410
Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2365. See equsalvw 1999 and equsalv 2250 for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsex 2411. (Contributed by NM, 2-Jun-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 5-Feb-2018.) (New usage is discouraged.)
Hypotheses
Ref Expression
equsal.1 𝑥𝜓
equsal.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
equsal (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)

Proof of Theorem equsal
StepHypRef Expression
1 equsal.1 . . 3 𝑥𝜓
2119.23 2196 . 2 (∀𝑥(𝑥 = 𝑦𝜓) ↔ (∃𝑥 𝑥 = 𝑦𝜓))
3 equsal.2 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
43pm5.74i 271 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦𝜓))
54albii 1813 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜓))
6 ax6e 2376 . . 3 𝑥 𝑥 = 𝑦
76a1bi 362 . 2 (𝜓 ↔ (∃𝑥 𝑥 = 𝑦𝜓))
82, 5, 73bitr4i 303 1 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wal 1531  wex 1773  wnf 1777
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-12 2163  ax-13 2365
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-nf 1778
This theorem is referenced by:  equsex  2411  equsalh  2413  dvelimf  2441  sb6x  2457  sb6rf  2461  bj-sbievv  36234
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