Theorem falnorfal 1599

Description: A ⊽ identity. (Contributed by Remi, 25-Oct-2023.) (Proof shortened by Wolf Lammen, 17-Dec-2023.)

Assertion

Ref	Expression
falnorfal	⊢ ((⊥ ⊽ ⊥) ↔ ⊤)

Proof of Theorem falnorfal

Step	Hyp	Ref	Expression
1		df-nor 1536	. . 3 ⊢ ((⊥ ⊽ ⊥) ↔ ¬ (⊥ ∨ ⊥))
2		falorfal 1587	. . 3 ⊢ ((⊥ ∨ ⊥) ↔ ⊥)
3	1, 2	xchbinx 335	. 2 ⊢ ((⊥ ⊽ ⊥) ↔ ¬ ⊥)
4		notfal 1575	. 2 ⊢ (¬ ⊥ ↔ ⊤)
5	3, 4	bitri 276	1 ⊢ ((⊥ ⊽ ⊥) ↔ ⊤)

Syntax hints:  ¬ wn 3   ↔ wb 207   ∨ wo 853   ⊽ wnor 1535  ⊤wtru 1548  ⊥wfal 1559

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 208  df-or 854  df-nor 1536  df-tru 1550  df-fal 1560

This theorem is referenced by: (None)