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Theorem nfbidf 2260
Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 4-Oct-2016.) df-nf 1816 changed. (Revised by Wolf Lammen, 18-Sep-2021.)
Hypotheses
Ref Expression
albid.1 Ⅎ𝑥𝜑
albid.2 (𝜑 → (𝜓 ↔ 𝜒))
Assertion
Ref Expression
nfbidf (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))

Proof of Theorem nfbidf
StepHypRef Expression
1 albid.1 . . . 4 Ⅎ𝑥𝜑
2 albid.2 . . . 4 (𝜑 → (𝜓 ↔ 𝜒))
31, 2exbid 2259 . . 3 (𝜑 → (∃𝑥𝜓 ↔ ∃𝑥𝜒))
41, 2albid 2258 . . 3 (𝜑 → (∀𝑥𝜓 ↔ ∀𝑥𝜒))
53, 4imbi12d 347 . 2 (𝜑 → ((∃𝑥𝜓 → ∀𝑥𝜓) ↔ (∃𝑥𝜒 → ∀𝑥𝜒)))
6 df-nf 1816 . 2 (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓))
7 df-nf 1816 . 2 (Ⅎ𝑥𝜒 ↔ (∃𝑥𝜒 → ∀𝑥𝜒))
85, 6, 73bitr4g 317 1 (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  ∃wex 1811  Ⅎwnf 1815
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212
This proof depends on definitions:  df-bi 210  df-ex 1812  df-nf 1816
This theorem is used by:  drnf2  2473  dvelimdf  2478  nfceqdf  2918  wl-nfimf1  38459  ichnfimlem  48545
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