Theorem trubifal 1406

Description: A ↔ identity. (Contributed by David A. Wheeler, 23-Feb-2018.)

Assertion

Ref	Expression
trubifal	⊢ ((⊤ ↔ ⊥) ↔ ⊥)

Proof of Theorem trubifal

Step	Hyp	Ref	Expression
1		dfbi2 386	. 2 ⊢ ((⊤ ↔ ⊥) ↔ ((⊤ → ⊥) ∧ (⊥ → ⊤)))
2		truimfal 1400	. . 3 ⊢ ((⊤ → ⊥) ↔ ⊥)
3		falimtru 1401	. . 3 ⊢ ((⊥ → ⊤) ↔ ⊤)
4	2, 3	anbi12i 456	. 2 ⊢ (((⊤ → ⊥) ∧ (⊥ → ⊤)) ↔ (⊥ ∧ ⊤))
5		falantru 1393	. 2 ⊢ ((⊥ ∧ ⊤) ↔ ⊥)
6	1, 4, 5	3bitri 205	1 ⊢ ((⊤ ↔ ⊥) ↔ ⊥)

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ⊤wtru 1344  ⊥wfal 1348

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-fal 1349

This theorem is referenced by:  falbitru  1407