Theorem nanbi1 1295

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 687	. . 3 ⊢ ((φ ↔ ψ) → ((φ ∧ χ) ↔ (ψ ∧ χ)))
2	1	notbid 285	. 2 ⊢ ((φ ↔ ψ) → (¬ (φ ∧ χ) ↔ ¬ (ψ ∧ χ)))
3		df-nan 1288	. 2 ⊢ ((φ ⊼ χ) ↔ ¬ (φ ∧ χ))
4		df-nan 1288	. 2 ⊢ ((ψ ⊼ χ) ↔ ¬ (ψ ∧ χ))
5	2, 3, 4	3bitr4g 279	1 ⊢ ((φ ↔ ψ) → ((φ ⊼ χ) ↔ (ψ ⊼ χ)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358   ⊼ wnan 1287

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

This proof depends on definitions:  df-bi 177  df-an 360  df-nan 1288

This theorem is used by:  nanbi2  1296  nanbi12  1297  nanbi1i  1298  nanbi1d  1301