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Theorem nfbid 1496
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, it is not free in (𝜓𝜒). (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 374 . 2 ((𝜓𝜒) ↔ ((𝜓𝜒) ∧ (𝜒𝜓)))
2 nfbid.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
3 nfbid.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
42, 3nfimd 1493 . . 3 (𝜑 → Ⅎ𝑥(𝜓𝜒))
53, 2nfimd 1493 . . 3 (𝜑 → Ⅎ𝑥(𝜒𝜓))
64, 5nfand 1476 . 2 (𝜑 → Ⅎ𝑥((𝜓𝜒) ∧ (𝜒𝜓)))
71, 6nfxfrd 1380 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wnf 1365
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-4 1416  ax-ial 1443  ax-i5r 1444
This theorem depends on definitions:  df-bi 114  df-nf 1366
This theorem is referenced by:  nfbi  1497  nfeudv  1931  nfeqd  2208  nfiotadxy  4897  iota2df  4918  bdsepnft  10373  strcollnft  10475
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