Theorem u1lemona 607

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemona	((a →1 b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))

Proof of Theorem u1lemona

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 37	. 2 ((a →1 b) ∪ a⊥ ) = ((a⊥ ∪ (a ∩ b)) ∪ a⊥ )
3		or32 75	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ a⊥ ) = ((a⊥ ∪ a⊥ ) ∪ (a ∩ b))
4		oridm 102	. . . 4 (a⊥ ∪ a⊥ ) = a⊥
5	4	ax-r5 37	. . 3 ((a⊥ ∪ a⊥ ) ∪ (a ∩ b)) = (a⊥ ∪ (a ∩ b))
6	3, 5	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
7	2, 6	ax-r2 35	1 ((a →1 b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  u1lemnaa 622  u1lem4 739

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-i1 43

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