Theorem bibi1i 227

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 139	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi.a	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 226	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 139	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 205	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 104

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  bibi12i  228  biadani  584  bilukdc  1357  sbrbis  1910  necon1abiddc  2344  necon1bbiddc  2345  necon4abiddc  2355  elrab3t  2808  ddifstab  3174  ssequn1  3212  asymref  4882