Theorem syl6rbb 253

Description: A syllogism inference from two biconditionals. (Contributed by NM, 5-Aug-1993.)

Hypotheses

Ref	Expression
syl6rbb.1	⊢ (φ → (ψ ↔ χ))
syl6rbb.2	⊢ (χ ↔ θ)

Assertion

Ref	Expression
syl6rbb	⊢ (φ → (θ ↔ ψ))

Proof of Theorem syl6rbb

Step	Hyp	Ref	Expression
1		syl6rbb.1	. . 3 ⊢ (φ → (ψ ↔ χ))
2		syl6rbb.2	. . 3 ⊢ (χ ↔ θ)
3	1, 2	syl6bb 252	. 2 ⊢ (φ → (ψ ↔ θ))
4	3	bicomd 192	1 ⊢ (φ → (θ ↔ ψ))

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

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

This theorem is used by:  syl6rbbr  255  bibif  335  pm5.61  693  oranabs  829  necon4bid  2583  resopab2  5002  funconstss  5407  scancan  6332