Theorem nanbi1 1447

Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)

Assertion

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

Proof of Theorem nanbi1

Step	Hyp	Ref	Expression
1		anbi1 739	. . 3 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ∧ 𝜒) ↔ (𝜓 ∧ 𝜒)))
2	1	notbid 307	. 2 ⊢ ((𝜑 ↔ 𝜓) → (¬ (𝜑 ∧ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒)))
3		df-nan 1440	. 2 ⊢ ((𝜑 ⊼ 𝜒) ↔ ¬ (𝜑 ∧ 𝜒))
4		df-nan 1440	. 2 ⊢ ((𝜓 ⊼ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒))
5	2, 3, 4	3bitr4g 302	1 ⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∧ wa 383   ⊼ wnan 1439

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 196  df-an 385  df-nan 1440

This theorem is referenced by:  nanbi2  1448  nanbi12  1449  nanbi1i  1450  nanbi1d  1453  nabi1  31559