Theorem u5lem1n 743

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

Assertion

Ref	Expression
u5lem1n	((a →5 b) →5 a)⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem u5lem1n

Step	Hyp	Ref	Expression
1		u5lem1 738	. . 3 ((a →5 b) →5 a) = ((a ∪ b) ∩ (a ∪ b⊥ ))
2		ancom 74	. . . 4 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a ∪ b⊥ ) ∩ (a ∪ b))
3		df-a 40	. . . . 5 ((a ∪ b⊥ ) ∩ (a ∪ b)) = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥
4		anor2 89	. . . . . . . 8 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
5		anor3 90	. . . . . . . 8 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
6	4, 5	2or 72	. . . . . . 7 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )
7	6	ax-r4 37	. . . . . 6 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥
8	7	ax-r1 35	. . . . 5 ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
9	3, 8	ax-r2 36	. . . 4 ((a ∪ b⊥ ) ∩ (a ∪ b)) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
10	2, 9	ax-r2 36	. . 3 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
11	1, 10	ax-r2 36	. 2 ((a →5 b) →5 a) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
12	11	con2 67	1 ((a →5 b) →5 a)⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ 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-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-i5 48  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u5lem2  748