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Theorem equsalhw 2225
Description: Version of equsalh 2355 with a disjoint variable condition, which does not require ax-13 2301. (Contributed by NM, 29-Nov-2015.) (Proof shortened by Wolf Lammen, 8-Jul-2022.)
Hypotheses
Ref Expression
equsalhw.1 (𝜓 → ∀𝑥𝜓)
equsalhw.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
equsalhw (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem equsalhw
StepHypRef Expression
1 equsalhw.1 . . 3 (𝜓 → ∀𝑥𝜓)
21nf5i 2084 . 2 𝑥𝜓
3 equsalhw.2 . 2 (𝑥 = 𝑦 → (𝜑𝜓))
42, 3equsalv 2196 1 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wal 1505
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-10 2079  ax-12 2106
This theorem depends on definitions:  df-bi 199  df-ex 1743  df-nf 1747
This theorem is referenced by:  dvelimhw  2279
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