Theorem u4lemnoa 663

Description: Lemma for non-tollens implication study. (Contributed by NM, 16-Dec-1997.)

Assertion

Ref	Expression
u4lemnoa	((a →4 b)⊥ ∪ a) = ((a ∪ b) ∩ (a ∪ b⊥ ))

Proof of Theorem u4lemnoa

Step	Hyp	Ref	Expression
1		u4lemana 608	. . . 4 ((a →4 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
2		ax-a2 31	. . . . 5 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b))
3		anor3 90	. . . . . 6 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
4		anor2 89	. . . . . 6 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
5	3, 4	2or 72	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )
6	2, 5	ax-r2 36	. . . 4 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )
7	1, 6	ax-r2 36	. . 3 ((a →4 b) ∩ a⊥ ) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ )
8		anor1 88	. . 3 ((a →4 b) ∩ a⊥ ) = ((a →4 b)⊥ ∪ a)⊥
9		oran3 93	. . 3 ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ ) = ((a ∪ b) ∩ (a ∪ b⊥ ))⊥
10	7, 8, 9	3tr2 64	. 2 ((a →4 b)⊥ ∪ a)⊥ = ((a ∪ b) ∩ (a ∪ b⊥ ))⊥
11	10	con1 66	1 ((a →4 b)⊥ ∪ a) = ((a ∪ b) ∩ (a ∪ b⊥ ))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 15

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i4 47  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u4lem1  737