Theorem orordir 113

Description: Distribution of disjunction over disjunction.

Assertion

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

Proof of Theorem orordir

Step	Hyp	Ref	Expression
1		oridm 110	. . . 4 (c ∪ c) = c
2	1	ax-r1 35	. . 3 c = (c ∪ c)
3	2	lor 70	. 2 ((a ∪ b) ∪ c) = ((a ∪ b) ∪ (c ∪ c))
4		or4 84	. 2 ((a ∪ b) ∪ (c ∪ c)) = ((a ∪ c) ∪ (b ∪ c))
5	3, 4	ax-r2 36	1 ((a ∪ b) ∪ c) = ((a ∪ c) ∪ (b ∪ c))

Syntax hints:   = wb 1   ∪ wo 6

This theorem is referenced by:  leror 152  wql2lem2 289  wleror 393  ska2 432

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

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