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  699  con2bidc  876  con2biddc  881  pm5.17dc  905  bigolden  957  nbbndc  1405  bilukdc  1407  falbitru  1428  3impexpbicom  1449  exists1  2141  eqcom  2198  abeq1  2306  necon2abiddc  2433  necon2bbiddc  2434  necon4bbiddc  2441  ssequn1  3333  axpow3  4210  isocnv  5858  suplocsrlem  7875  uzennn  10528  bezoutlemle  12175