Theorem orordi 104

Description: Distribution of disjunction over disjunction.

Assertion

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

Proof of Theorem orordi

Step	Hyp	Ref	Expression
1		oridm 102	. . . 4 (a ∪ a) = a
2	1	ax-r1 34	. . 3 a = (a ∪ a)
3	2	ax-r5 37	. 2 (a ∪ (b ∪ c)) = ((a ∪ a) ∪ (b ∪ c))
4		or4 77	. 2 ((a ∪ a) ∪ (b ∪ c)) = ((a ∪ b) ∪ (a ∪ c))
5	3, 4	ax-r2 35	1 (a ∪ (b ∪ c)) = ((a ∪ b) ∪ (a ∪ c))

Syntax hints:   = wb 1   ∪ wo 6

This theorem is referenced by:  ska2 414  lem4 493  i3abs1 504  u12lem 753  orbi 824  i1orni1 829

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-t 40  df-f 41

metamath.org