ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  nfbid Structured version   GIF version

Theorem nfbid 1477
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 368 . 2 ((ψχ) ↔ ((ψχ) (χψ)))
2 nfbid.1 . . . 4 (φ → Ⅎxψ)
3 nfbid.2 . . . 4 (φ → Ⅎxχ)
42, 3nfimd 1474 . . 3 (φ → Ⅎx(ψχ))
53, 2nfimd 1474 . . 3 (φ → Ⅎx(χψ))
64, 5nfand 1457 . 2 (φ → Ⅎx((ψχ) (χψ)))
71, 6nfxfrd 1361 1 (φ → Ⅎx(ψχ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  wnf 1346
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 1333  ax-gen 1335  ax-4 1397  ax-ial 1424  ax-i5r 1425
This theorem depends on definitions:  df-bi 110  df-nf 1347
This theorem is referenced by:  nfbi  1478  nfeudv  1912  nfeqd  2189  nfiotadxy  4813  iota2df  4834  bdsepnft  9321  strcollnft  9414
  Copyright terms: Public domain W3C validator