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Theorem nfbi 1569
 Description: If 𝑥 is not free in 𝜑 and 𝜓, then it is not free in (𝜑 ↔ 𝜓). (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 2-Jan-2018.)
Hypotheses
Ref Expression
nfbi.1 𝑥𝜑
nfbi.2 𝑥𝜓
Assertion
Ref Expression
nfbi 𝑥(𝜑𝜓)

Proof of Theorem nfbi
StepHypRef Expression
1 nfbi.1 . . . 4 𝑥𝜑
21a1i 9 . . 3 (⊤ → Ⅎ𝑥𝜑)
3 nfbi.2 . . . 4 𝑥𝜓
43a1i 9 . . 3 (⊤ → Ⅎ𝑥𝜓)
52, 4nfbid 1568 . 2 (⊤ → Ⅎ𝑥(𝜑𝜓))
65mptru 1341 1 𝑥(𝜑𝜓)
 Colors of variables: wff set class Syntax hints:   ↔ wb 104  ⊤wtru 1333  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-tru 1335  df-nf 1438 This theorem is referenced by:  sb8eu  2013  nfeuv  2018  bm1.1  2125  abbi  2254  nfeq  2290  cleqf  2306  sbhypf  2738  ceqsexg  2816  elabgt  2828  elabgf  2829  copsex2t  4174  copsex2g  4175  opelopabsb  4189  opeliunxp2  4686  ralxpf  4692  rexxpf  4693  cbviota  5100  sb8iota  5102  fmptco  5593  nfiso  5714  dfoprab4f  6098  opeliunxp2f  6142  xpf1o  6745  bdsepnfALT  13256
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