Theorem nanbi1 1500

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 347	. 2 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 → ¬ 𝜒) ↔ (𝜓 → ¬ 𝜒)))
2		dfnan2 1493	. 2 ⊢ ((𝜑 ⊼ 𝜒) ↔ (𝜑 → ¬ 𝜒))
3		dfnan2 1493	. 2 ⊢ ((𝜓 ⊼ 𝜒) ↔ (𝜓 → ¬ 𝜒))
4	1, 2, 3	3bitr4g 314	1 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ⊼ wnan 1490

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-an 396  df-nan 1491

This theorem is referenced by:  nanbi2  1501  nanbi12  1502  nanbi1i  1503  nanbi1d  1506