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Theorem nfbid 1568
 Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then 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 386 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1565 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1565 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 1548 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1452 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  Ⅎwnf 1437 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 1424  ax-gen 1426  ax-4 1488  ax-ial 1515  ax-i5r 1516 This theorem depends on definitions:  df-bi 116  df-nf 1438 This theorem is referenced by:  nfbi  1569  nfeudv  2015  nfeqd  2297  nfiotadw  5100  iota2df  5121  bdsepnft  13276
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