Theorem u1lemona 625

Description: Lemma for Sasaki implication study. (Contributed by NM, 15-Dec-1997.)

Assertion

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

Proof of Theorem u1lemona

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 38	. 2 ((a →1 b) ∪ a⊥ ) = ((a⊥ ∪ (a ∩ b)) ∪ a⊥ )
3		or32 82	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ a⊥ ) = ((a⊥ ∪ a⊥ ) ∪ (a ∩ b))
4		oridm 110	. . . 4 (a⊥ ∪ a⊥ ) = a⊥
5	4	ax-r5 38	. . 3 ((a⊥ ∪ a⊥ ) ∪ (a ∩ b)) = (a⊥ ∪ (a ∩ b))
6	3, 5	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
7	2, 6	ax-r2 36	1 ((a →1 b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This proof depends on 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 proof depends on definitions:  df-t 41  df-f 42  df-i1 44

This theorem is used by:  u1lemnaa  640  u1lem4  757