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Theorem nfbid 1462
 Description: If in a context x is not free in ψ and χ, it is not free in (ψ ↔ χ). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.)
Hypotheses
Ref Expression
nfbid.1 (φ → Ⅎxψ)
nfbid.2 (φ → Ⅎxχ)
Assertion
Ref Expression
nfbid (φ → Ⅎx(ψχ))

Proof of Theorem nfbid
StepHypRef Expression
1 dfbi2 368 . 2 ((ψχ) ↔ ((ψχ) (χψ)))
2 nfbid.1 . . . 4 (φ → Ⅎxψ)
3 nfbid.2 . . . 4 (φ → Ⅎxχ)
42, 3nfimd 1459 . . 3 (φ → Ⅎx(ψχ))
53, 2nfimd 1459 . . 3 (φ → Ⅎx(χψ))
64, 5nfand 1442 . 2 (φ → Ⅎx((ψχ) (χψ)))
71, 6nfxfrd 1344 1 (φ → Ⅎx(ψχ))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  Ⅎwnf 1329 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1316  ax-gen 1318  ax-4 1381  ax-ial 1409  ax-i5r 1410 This theorem depends on definitions:  df-bi 110  df-nf 1330 This theorem is referenced by:  nfbi  1463  nfeudv  1897  nfeqd  2174  nfiotadxy  4797  iota2df  4818  bdsepnft  7260  strcollnft  7349
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