Theorem nanbi12d 1538

Description: Join two logical equivalences with anti-conjunction. (Contributed by Scott Fenton, 2-Jan-2018.)

Hypotheses

Ref	Expression
nanbid.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))
nanbi12d.2	⊢ (𝜑 → (𝜃 ↔ 𝜏))

Assertion

Ref	Expression
nanbi12d	⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜏)))

Proof of Theorem nanbi12d

Step	Hyp	Ref	Expression
1		nanbid.1	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2		nanbi12d.2	. 2 ⊢ (𝜑 → (𝜃 ↔ 𝜏))
3		nanbi12 1532	. 2 ⊢ (((𝜓 ↔ 𝜒) ∧ (𝜃 ↔ 𝜏)) → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜏)))
4	1, 2, 3	syl2anc 596	1 ⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜏)))

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:  nanass  1539