Theorem bibi1i 338

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 222	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi2i.1	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 337	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 222	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 297	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 206

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 207

This theorem is referenced by:  bibi12i  339  biluk  385  biadaniALT  820  nanass  1511  xorass  1516  hadbi  1599  hadcoma  1600  hadnot  1603  sbrbis  2315  csbied  3885  dfss2  3919  ssequn1  4138  asymref  6073  aceq1  10027  aceq0  10028  zfac  10370  zfcndac  10530  hashreprin  34777  axacprim  35901  eliminable-abeqv  37068  wl-3xorcoma  37683  wl-3xornot  37686  redundpbi1  38888  onsupmaxb  43481  rp-fakeanorass  43754  ichn  47702  dfich2  47704