Theorem nanbi1 1492

Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by Anthony Hart, 1-Sep-2011.) (Proof shortened by Wolf Lammen, 27-Jun-2020.)

Assertion

Ref	Expression
nanbi1	⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒)))

Proof of Theorem nanbi1

Step	Hyp	Ref	Expression
1		imbi1 351	. 2 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 → ¬ 𝜒) ↔ (𝜓 → ¬ 𝜒)))
2		nanimn 1485	. 2 ⊢ ((𝜑 ⊼ 𝜒) ↔ (𝜑 → ¬ 𝜒))
3		nanimn 1485	. 2 ⊢ ((𝜓 ⊼ 𝜒) ↔ (𝜓 → ¬ 𝜒))
4	1, 2, 3	3bitr4g 317	1 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ⊼ wnan 1482

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 210  df-an 400  df-nan 1483

This theorem is referenced by:  nanbi2  1493  nanbi12  1494  nanbi1i  1495  nanbi1d  1498