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Theorem bj-nfn 31795
 Description: A variable is non-free in a proposition if and only if it is so in its negation. Requires fewer axioms than nfn 1768. (Contributed by BJ, 6-May-2019.)
Assertion
Ref Expression
bj-nfn (ℲℲ𝑥𝜑 ↔ ℲℲ𝑥 ¬ 𝜑)

Proof of Theorem bj-nfn
StepHypRef Expression
1 notnotb 303 . . . . 5 (𝜑 ↔ ¬ ¬ 𝜑)
21albii 1737 . . . 4 (∀𝑥𝜑 ↔ ∀𝑥 ¬ ¬ 𝜑)
32orbi1i 541 . . 3 ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ ¬ 𝜑 ∨ ∀𝑥 ¬ 𝜑))
4 orcom 401 . . 3 ((∀𝑥 ¬ ¬ 𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
53, 4bitri 263 . 2 ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
6 bj-nf3 31767 . 2 (ℲℲ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
7 bj-nf3 31767 . 2 (ℲℲ𝑥 ¬ 𝜑 ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
85, 6, 73bitr4i 291 1 (ℲℲ𝑥𝜑 ↔ ℲℲ𝑥 ¬ 𝜑)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 195   ∨ wo 382  ∀wal 1473  ℲℲwnff 31764 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728 This theorem depends on definitions:  df-bi 196  df-or 384  df-ex 1696  df-bj-nf 31765 This theorem is referenced by: (None)
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