Theorem syl5rbb 411

Description: A syllogism inference from two biconditionals.

Hypotheses

Ref	Expression
syl5rbb.1	⊢ (φ → (ψ ↔ χ))
syl5rbb.2	⊢ (θ ↔ ψ)

Assertion

Ref	Expression
syl5rbb	⊢ (φ → (χ ↔ θ))

Proof of Theorem syl5rbb

Step	Hyp	Ref	Expression
1		syl5rbb.1	. . 3 ⊢ (φ → (ψ ↔ χ))
2		syl5rbb.2	. . 3 ⊢ (θ ↔ ψ)
3	1, 2	syl5bb 410	. 2 ⊢ (φ → (θ ↔ χ))
4	3	bicomd 399	1 ⊢ (φ → (χ ↔ θ))

Syntax hints:   → wi 2   ↔ wb 127

This theorem is referenced by:  syl5rbbr 413  fnresdisj 2732  f1oiso 2942  rdglim2 2987  1idpr 3927

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198

metamath.org