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Theorem nfbid 1832
 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 609 . 2 ((ψχ) ↔ ((ψχ) (χψ)))
2 nfbid.1 . . . 4 (φ → Ⅎxψ)
3 nfbid.2 . . . 4 (φ → Ⅎxχ)
42, 3nfimd 1808 . . 3 (φ → Ⅎx(ψχ))
53, 2nfimd 1808 . . 3 (φ → Ⅎx(χψ))
64, 5nfand 1822 . 2 (φ → Ⅎx((ψχ) (χψ)))
71, 6nfxfrd 1571 1 (φ → Ⅎx(ψχ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  Ⅎwnf 1544 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746 This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-nf 1545 This theorem is referenced by:  nfbi  1834  nfeud2  2216  nfeqd  2503  nfiotad  4342  iota2df  4365
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