Theorem bicom 140

Description: Commutative law for equivalence. Theorem *4.21 of [WhiteheadRussell] p. 117. (Contributed by NM, 5-Aug-1993.) (Revised by NM, 11-Nov-2012.)

Assertion

Ref	Expression
bicom	⊢ ((𝜑 ↔ 𝜓) ↔ (𝜓 ↔ 𝜑))

Proof of Theorem bicom

Step	Hyp	Ref	Expression
1		bicom1 131	. 2 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 ↔ 𝜑))
2		bicom1 131	. 2 ⊢ ((𝜓 ↔ 𝜑) → (𝜑 ↔ 𝜓))
3	1, 2	impbii 126	1 ⊢ ((𝜑 ↔ 𝜓) ↔ (𝜓 ↔ 𝜑))

Syntax hints:   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  bicomd  141  bibi1i  228  bibi1d  233  ibibr  246  bibif  706  con2bidc  883  con2biddc  888  pm5.17dc  912  bigolden  964  nbbndc  1439  bilukdc  1441  falbitru  1462  3impexpbicom  1484  exists1  2179  eqcom  2236  abeq1  2344  eqabcb  2371  necon2abiddc  2480  necon2bbiddc  2481  necon4bbiddc  2488  ssequn1  3393  axpow3  4295  isocnv  5990  suplocsrlem  8139  uzennn  10822  bezoutlemle  12729