Theorem nanbi1d 1301

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 1295	. 2 ⊢ ((ψ ↔ χ) → ((ψ ⊼ θ) ↔ (χ ⊼ θ)))
3	1, 2	syl 15	1 ⊢ (φ → ((ψ ⊼ θ) ↔ (χ ⊼ θ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ⊼ 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:  nineq1  3235