Theorem bibi1i 328

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 212	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi2i.1	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 327	. 2 ⊢ ((𝜒 ↔ 𝜑) ↔ (𝜒 ↔ 𝜓))
4		bicom 212	. 2 ⊢ ((𝜒 ↔ 𝜓) ↔ (𝜓 ↔ 𝜒))
5	1, 3, 4	3bitri 286	1 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 196

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 197

This theorem is referenced by:  bibi12i  329  biluk  974  xorass  1468  hadbi  1537  hadnot  1541  sbrbis  2405  ssequn1  3783  symdifass  3853  asymref  5512  aceq1  8940  aceq0  8941  zfac  9282  zfcndac  9441  funcnvmptOLD  29467  hashreprin  30698  axacprim  31584  rp-fakeanorass  37858  rp-fakenanass  37860