Theorem bibi1i 327

Description: Inference adding a biconditional to the right in an equivalence. (Contributed by NM, 26-May-1993.)

Hypothesis

Ref	Expression
bibi2i.1	⊢ (𝜑 ↔ 𝜓)

Assertion

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

Proof of Theorem bibi1i

Step	Hyp	Ref	Expression
1		bicom 211	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi2i.1	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 326	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 211	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 285	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 195

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 196

This theorem is referenced by:  bibi12i  328  biluk  970  xorass  1460  hadbi  1528  hadnot  1532  sbrbis  2393  ssequn1  3745  symdifass  3815  asymref  5431  aceq1  8823  aceq0  8824  zfac  9165  zfcndac  9320  funcnvmptOLD  28850  axacprim  30838  rp-fakeanorass  36877  rp-fakenanass  36879