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

Step	Hyp	Ref	Expression
1		notnotb 303	. . . . 5 ⊢ (𝜑 ↔ ¬ ¬ 𝜑)
2	1	albii 1737	. . . 4 ⊢ (∀𝑥𝜑 ↔ ∀𝑥 ¬ ¬ 𝜑)
3	2	orbi1i 541	. . 3 ⊢ ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ ¬ 𝜑 ∨ ∀𝑥 ¬ 𝜑))
4		orcom 401	. . 3 ⊢ ((∀𝑥 ¬ ¬ 𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
5	3, 4	bitri 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))
8	5, 6, 7	3bitr4i 291	1 ⊢ (ℲℲ𝑥𝜑 ↔ ℲℲ𝑥 ¬ 𝜑)

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)