Theorem nanbi1d 1536

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

Hypothesis

Ref	Expression
nanbid.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
nanbi1d	⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜃)))

Proof of Theorem nanbi1d

Step	Hyp	Ref	Expression
1		nanbid.1	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2		nanbi1 1530	. 2 ⊢ ((𝜓 ↔ 𝜒) → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜃)))
3	1, 2	syl 18	1 ⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜃)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ⊼ wnan 1520

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-nan 1521

This theorem is used by: (None)