Theorem syl6rbb 195

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 194	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜃))
4	3	bicomd 139	1 ⊢ (𝜑 → (𝜃 ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  syl6rbbr  197  bibif  646  pm5.61  740  oranabs  761  pm5.7dc  895  nbbndc  1325  resopab2  4675  xpcom  4884  f1od2  5876  ac6sfi  6379  elznn0  8366  rexuz3  9876