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  2177  eqcom  2234  abeq1  2342  eqabcb  2369  necon2abiddc  2478  necon2bbiddc  2479  necon4bbiddc  2486  ssequn1  3389  axpow3  4290  isocnv  5984  suplocsrlem  8123  uzennn  10798  bezoutlemle  12704