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Theorem nf4r 1577
Description: If 𝜑 is always true or always false, then variable 𝑥 is effectively not free in 𝜑. The converse holds given a decidability condition, as seen at nf4dc 1576. (Contributed by Jim Kingdon, 21-Jul-2018.)
Assertion
Ref Expression
nf4r ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) → Ⅎ𝑥𝜑)

Proof of Theorem nf4r
StepHypRef Expression
1 orcom 657 . . 3 ((¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑) ↔ (∀𝑥𝜑 ∨ ¬ ∃𝑥𝜑))
2 alnex 1404 . . . 4 (∀𝑥 ¬ 𝜑 ↔ ¬ ∃𝑥𝜑)
32orbi2i 689 . . 3 ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥𝜑 ∨ ¬ ∃𝑥𝜑))
41, 3bitr4i 180 . 2 ((¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑) ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
5 imorr 808 . . 3 ((¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑) → (∃𝑥𝜑 → ∀𝑥𝜑))
6 nf2 1574 . . 3 (Ⅎ𝑥𝜑 ↔ (∃𝑥𝜑 → ∀𝑥𝜑))
75, 6sylibr 141 . 2 ((¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑) → Ⅎ𝑥𝜑)
84, 7sylbir 129 1 ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) → Ⅎ𝑥𝜑)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wo 639  wal 1257  wnf 1365  wex 1397
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-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-gen 1354  ax-ie2 1399  ax-4 1416  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-tru 1262  df-fal 1265  df-nf 1366
This theorem is referenced by: (None)
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