Theorem bibi1i 341

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 225	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi2i.1	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 340	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 225	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 300	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

This proof depends on syntax axioms:   ↔ wb 209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210

This theorem is used by:  bibi12i  342  biluk  390  biadaniALT  833  nanass  1539  xorass  1544  hadbi  1627  hadcoma  1628  hadnot  1631  sbrbis  2342  csbied  3882  dfss2  4094  ssequn1  4184  asymref  6109  aceq1  10199  aceq0  10200  zfac  10541  zfcndac  10707  hashreprin  35295  axacprim  36564  eliminable-abeqv  37871  wl-3xorcoma  38494  wl-3xornot  38497  redundpbi1  39740  onsupmaxb  44334  rp-fakeanorass  44607  ichn  48631  dfich2  48633