Theorem u2lemob 613

Description: Lemma for Dishkant implication study.

Assertion

Ref	Expression
u2lemob	((a →2 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)

Proof of Theorem u2lemob

Step	Hyp	Ref	Expression
1		df-i2 44	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2	1	ax-r5 37	. 2 ((a →2 b) ∪ b) = ((b ∪ (a⊥ ∩ b⊥ )) ∪ b)
3		or32 75	. . 3 ((b ∪ (a⊥ ∩ b⊥ )) ∪ b) = ((b ∪ b) ∪ (a⊥ ∩ b⊥ ))
4		ax-a2 30	. . . 4 ((b ∪ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (b ∪ b))
5		oridm 102	. . . . 5 (b ∪ b) = b
6	5	lor 66	. . . 4 ((a⊥ ∩ b⊥ ) ∪ (b ∪ b)) = ((a⊥ ∩ b⊥ ) ∪ b)
7	4, 6	ax-r2 35	. . 3 ((b ∪ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ b)
8	3, 7	ax-r2 35	. 2 ((b ∪ (a⊥ ∩ b⊥ )) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)
9	2, 8	ax-r2 35	1 ((a →2 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 14

This theorem is referenced by:  u2lemnanb 638

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  df-i2 44

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