 Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  equsalvw Structured version   Visualization version   GIF version

Theorem equsalvw 1918
 Description: Version of equsal 2279 with two dv conditions, which requires fewer axioms. (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 19.23v 1889 . 2 (∀𝑥(𝑥 = 𝑦𝜓) ↔ (∃𝑥 𝑥 = 𝑦𝜓))
2 equsalvw.1 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
32pm5.74i 259 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦𝜓))
43albii 1737 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜓))
5 ax6ev 1877 . . 3 𝑥 𝑥 = 𝑦
65a1bi 351 . 2 (𝜓 ↔ (∃𝑥 𝑥 = 𝑦𝜓))
71, 4, 63bitr4i 291 1 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473  ∃wex 1695 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875 This theorem depends on definitions:  df-bi 196  df-ex 1696 This theorem is referenced by:  ax13lem2  2284  reu8  3369  asymref2  5432  intirr  5433  fun11  5877  bj-dvelimdv  32027  bj-dvelimdv1  32028  undmrnresiss  36929  pm13.192  37633
 Copyright terms: Public domain W3C validator