Theorem bibi1i 228

Description: Inference adding a biconditional to the right in an equivalence. (Contributed by NM, 5-Aug-1993.)

Hypothesis

Ref	Expression
bibi.a	⊢ (𝜑 ↔ 𝜓)

Assertion

Ref	Expression
bibi1i	⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Proof of Theorem bibi1i

Step	Hyp	Ref	Expression
1		bicom 140	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi.a	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 227	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 140	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 206	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  bibi12i  229  biadani  612  bilukdc  1396  sbrbis  1961  necon1abiddc  2409  necon1bbiddc  2410  necon4abiddc  2420  elrab3t  2894  ddifstab  3269  ssequn1  3307  asymref  5016