Theorem u5lemob 634

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

Assertion

Ref	Expression
u5lemob	((a →5 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)

Proof of Theorem u5lemob

Step	Hyp	Ref	Expression
1		df-i5 48	. . 3 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
2	1	ax-r5 38	. 2 ((a →5 b) ∪ b) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b)
3		ax-a3 32	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∪ b))
4		lear 161	. . . . . 6 (a ∩ b) ≤ b
5		lear 161	. . . . . 6 (a⊥ ∩ b) ≤ b
6	4, 5	lel2or 170	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) ≤ b
7		leor 159	. . . . 5 b ≤ ((a⊥ ∩ b⊥ ) ∪ b)
8	6, 7	letr 137	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b)) ≤ ((a⊥ ∩ b⊥ ) ∪ b)
9	8	df-le2 131	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∪ b)) = ((a⊥ ∩ b⊥ ) ∪ b)
10	3, 9	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)
11	2, 10	ax-r2 36	1 ((a →5 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 16

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-a 40  df-t 41  df-f 42  df-i5 48  df-le1 130  df-le2 131

This theorem is used by:  u5lemnanb  659