Theorem bisym1 36420

Description: A symmetry with ↔.

See negsym1 36418 for more information. (Contributed by Anthony Hart, 4-Sep-2011.)

Assertion

Ref	Expression
bisym1	⊢ ((𝜓 ↔ (𝜓 ↔ ⊥)) → (𝜓 ↔ 𝜑))

Proof of Theorem bisym1

Step	Hyp	Ref	Expression
1		nbfal 1555	. . 3 ⊢ (¬ 𝜓 ↔ (𝜓 ↔ ⊥))
2	1	bibi2i 337	. 2 ⊢ ((𝜓 ↔ ¬ 𝜓) ↔ (𝜓 ↔ (𝜓 ↔ ⊥)))
3		pm5.19 386	. . 3 ⊢ ¬ (𝜓 ↔ ¬ 𝜓)
4	3	pm2.21i 119	. 2 ⊢ ((𝜓 ↔ ¬ 𝜓) → (𝜓 ↔ 𝜑))
5	2, 4	sylbir 235	1 ⊢ ((𝜓 ↔ (𝜓 ↔ ⊥)) → (𝜓 ↔ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206  ⊥wfal 1552

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 207  df-tru 1543  df-fal 1553

This theorem is referenced by: (None)