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Theorem nfbidf 1502
Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 4-Oct-2016.)
Hypotheses
Ref Expression
nfbidf.1 𝑥𝜑
nfbidf.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
nfbidf (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))

Proof of Theorem nfbidf
StepHypRef Expression
1 nfbidf.1 . . . 4 𝑥𝜑
21nfri 1482 . . 3 (𝜑 → ∀𝑥𝜑)
3 nfbidf.2 . . . 4 (𝜑 → (𝜓𝜒))
42, 3albidh 1439 . . . 4 (𝜑 → (∀𝑥𝜓 ↔ ∀𝑥𝜒))
53, 4imbi12d 233 . . 3 (𝜑 → ((𝜓 → ∀𝑥𝜓) ↔ (𝜒 → ∀𝑥𝜒)))
62, 5albidh 1439 . 2 (𝜑 → (∀𝑥(𝜓 → ∀𝑥𝜓) ↔ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7 df-nf 1420 . 2 (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
8 df-nf 1420 . 2 (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
96, 7, 83bitr4g 222 1 (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wal 1312  wnf 1419
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-4 1470
This theorem depends on definitions:  df-bi 116  df-nf 1420
This theorem is referenced by:  dvelimdf  1967  nfcjust  2244  nfceqdf  2255
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