Theorem orordi 112

Description: Distribution of disjunction over disjunction. (Contributed by NM, 27-Aug-1997.)

Assertion

Ref	Expression
orordi	(a ∪ (b ∪ c)) = ((a ∪ b) ∪ (a ∪ c))

Proof of Theorem orordi

Step	Hyp	Ref	Expression
1		oridm 110	. . . 4 (a ∪ a) = a
2	1	ax-r1 35	. . 3 a = (a ∪ a)
3	2	ax-r5 38	. 2 (a ∪ (b ∪ c)) = ((a ∪ a) ∪ (b ∪ c))
4		or4 84	. 2 ((a ∪ a) ∪ (b ∪ c)) = ((a ∪ b) ∪ (a ∪ c))
5	3, 4	ax-r2 36	1 (a ∪ (b ∪ c)) = ((a ∪ b) ∪ (a ∪ c))

Syntax hints:   = wb 1   ∪ wo 6

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-t 41  df-f 42

This theorem is referenced by:  ska2  432  lem4  511  i3abs1  522  u12lem  771  orbi  842  i1orni1  847  lem4.6.6i1j3  1094