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 212	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜒 ↔ 𝜑))
2		bibi2i.1	. . 3 ⊢ (𝜑 ↔ 𝜓)
3	2	bibi2i 326	. 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  328  biluk  994  xorass  1508  hadbi  1577  hadnot  1581  sbrbis  2433  ssequn1  3816  symdifass  3886  asymref  5547  aceq1  8978  aceq0  8979  zfac  9320  zfcndac  9479  funcnvmptOLD  29595  hashreprin  30826  axacprim  31710  rp-fakeanorass  38175  rp-fakenanass  38177