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Theorem nfnbi 1856
Description: A variable is non-free in a proposition if and only if it is so in its negation. (Contributed by BJ, 6-May-2019.)
Assertion
Ref Expression
nfnbi (Ⅎ𝑥𝜑 ↔ Ⅎ𝑥 ¬ 𝜑)

Proof of Theorem nfnbi
StepHypRef Expression
1 orcom 867 . 2 ((∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑) ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥𝜑))
2 nf3 1788 . 2 (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ∀𝑥 ¬ 𝜑))
3 nf3 1788 . . 3 (Ⅎ𝑥 ¬ 𝜑 ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
4 notnotb 318 . . . . 5 (𝜑 ↔ ¬ ¬ 𝜑)
54albii 1821 . . . 4 (∀𝑥𝜑 ↔ ∀𝑥 ¬ ¬ 𝜑)
65orbi2i 910 . . 3 ((∀𝑥 ¬ 𝜑 ∨ ∀𝑥𝜑) ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥 ¬ ¬ 𝜑))
73, 6bitr4i 281 . 2 (Ⅎ𝑥 ¬ 𝜑 ↔ (∀𝑥 ¬ 𝜑 ∨ ∀𝑥𝜑))
81, 2, 73bitr4i 306 1 (Ⅎ𝑥𝜑 ↔ Ⅎ𝑥 ¬ 𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 209  wo 844  wal 1536  wnf 1785
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811
This theorem depends on definitions:  df-bi 210  df-or 845  df-ex 1782  df-nf 1786
This theorem is referenced by:  nfnt  1857  wl-sb8et  34953
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