Theorem bicom 96

Description: Commutative law. (Contributed by NM, 10-Aug-1997.)

Assertion

Ref	Expression
bicom	(a ≡ b) = (b ≡ a)

Proof of Theorem bicom

Step	Hyp	Ref	Expression
1		ancom 74	. . 3 (a ∩ b) = (b ∩ a)
2		ancom 74	. . 3 (a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ )
3	1, 2	2or 72	. 2 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((b ∩ a) ∪ (b⊥ ∩ a⊥ ))
4		dfb 94	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
5		dfb 94	. 2 (b ≡ a) = ((b ∩ a) ∪ (b⊥ ∩ a⊥ ))
6	3, 4, 5	3tr1 63	1 (a ≡ b) = (b ≡ a)

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-b 39  df-a 40

This theorem is referenced by:  rbi  98  wr1  197  wwfh1  216  wwfh2  217  ska12  240  nomcon5  306  nom35  324  nom55  336  nom65  342  ka4ot  435  ublemc2  729  mlaconj4  844  distid  887  oago3.29  889  oago3.21x  890