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Theorem equsalvw 2006
Description: Version of equsalv 2264 with a disjoint variable condition, and of equsal 2435 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2007. (Contributed by BJ, 31-May-2019.)
Hypothesis
Ref Expression
equsalvw.1 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
equsalvw (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Distinct variable groups:   𝑥,𝑦   𝜓,𝑥
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑦)

Proof of Theorem equsalvw
StepHypRef Expression
1 equsalvw.1 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
21pm5.74i 273 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦𝜓))
32albii 1816 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜓))
4 equsv 2005 . 2 (∀𝑥(𝑥 = 𝑦𝜓) ↔ 𝜓)
53, 4bitri 277 1 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wal 1531
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966
This theorem depends on definitions:  df-bi 209  df-ex 1777
This theorem is referenced by:  equsexvw  2007  equvelv  2034  sb6  2089  sbievw  2099  ax13lem2  2390  reu8  3723  asymref2  5976  intirr  5977  fun11  6427  fv3  6687  fpwwe2lem12  10062  bj-dvelimdv  34175  bj-dvelimdv1  34176  wl-dfralflem  34837  undmrnresiss  39964  pm13.192  40742
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